mm 


. I ^  J.. --^— -^.—^ «-.-  ^« 


^*i 


Digitized  by  the  Internet  Archive 

in  2007  with  funding  from 

IVIicrosoft  Corporation 


http://www.archive.org/details/elementsofplanegOOsandrich 


ELEMENTS 


OF 


PLANE     GEOMETRY 


BY 

ALAN    SANDERS 

HUGHES   HIGH   SCHOOL,    CINCINNATI,   OHIO 


>>•?€ 


NEW   YORK . : .  CINCINNATI  • : .  CHICAGO 

AMERICAN    BOOK    COMPANY 


COPYEIGHT,   1901,   BY 

AMEEICAN  BOOK   COMPANY. 

Entered  at  Stationers'  Hall,  London. 


BANDBES'  PLANE  GEOM. 

E-P    2 


PURPOSE   AND   DISTINCTIVE   FEATURES 

This  work  has  been  prepared  for  the  use  of  classes  in 
high  schools,  academies,  and  preparatory  schools.  Its 
distinctive  features  are  :  — 

1.  The  omission  of  parts  of  demonstrations. 

By  this  expedient  the  student  is  forced  to  rely  more 
on  his  own  reasoning  powers,  and  is  prevented  from 
acquiring  the  detrimental  habit  of  memorizing  the  text. 

As  it  is  necessary  for  the  beginner  in  Geometry  to 
learn  the  form  of  a  geometrical  demonstration,  the  demon- 
strations of  the  first  few  propositions  are  given  in  full. 
In  the  succeeding  propositions  only  the  most  obvious 
steps  are  omitted,  the  omission  in  each  case  being  indi- 
cated by  an  interrogation  mark  (?).  In  no  case  is  the 
student  expected  to  originate  the  plan  of  proof. 

2.  The  introduction^  after  each  proposition,  of  exercises 
bearing  directly  upon  the  principle  of  the  proposition. 

As  soon  as  a  proposition  has  been  mastered,  the  stu- 
dent is  required  to  apply  its  principle  in  the  solution 
of  a  series  of  easy  exercises.  Hints  or  suggestions  are 
given   to   aid    the    pupil    in   the    solution    of    the    more 

difficult  exercises. 

3 


4  PURPOSE   AND   DISTINCTIVE    FEATURES 

8.  All  constructions^  such  as  draiving  parallels^  erecting 
perpendiculars,  etc.,  are  given  before  they  are  required  to 
he  used  in  demonstrations. 

4.  Exercises  in  Modern  Greometry, 

Exercises  involving  the  principles  of  Modern  Geometry 
are  .given  under  their  proper  propositions.  As  the  omis- 
sion of  these  exercises  cannot  affect  the  sequence  of 
propositions,  they  may  be  disregarded  at  the  discretion 
of  the  teacher. 

5.  Propositions  and  converses. 

Whenever  possible,  the  converse  of  a  proposition  is 
given  with  the  proposition  itself. 

6.  Number  of  exercises. 

Besides  the  exercises  directly  following  each  proposi- 
tion, miscellaneous  exercises  are  given  at  the  end  of  each 
book.  It  may  be  found  that  there  are  more  exercises 
given  than  can  be  covered  by  a  class  in  the  time  allotted 
to  the  subject  of  Plane  Geometry ;  in  which  case  the 
teacher  will  have  to  select  from  the  lists  given. 

While  the  exercises  have  been  drawn  from  many 
sources,  the  author  has  availed  himself  in  particular  of 
the  recent  entrance  examination  papers  of  the  best  Ameri- 
can colleges  and  scientific  schools. 

The  author  wishes  to  express  his  obligations  to  his 
colleagues  in  the  Cincinnati  Higli  Schools  for  their  criti- 
cism and  encouragement,  and  especially  to  Miss  Celia 
Doerner  of  Hughes  High  School  for  valuable  suggestions 
and  for  her  painstaking  reading  of  the  proof. 


CONTENTS 

PAGE 

Preliminary  Definitions .9 

Axioms 12 

Postulates 13 

Symbols  and  Abbreviations ,        .  14 

BOOK   I 

Rectilinear  Figures 15 

Additional  Exercises 71 

BOOK  IT 

Circles 79 

Additional  Exercises ,        .        .  122 

BOOK  III 

Proportions  —  Similar  Polygons    ......  129 

Additional  Exercises <,  173 

BOOK   IV 

Areas  of  Polygons 181 

Additional  Exercises »  213 

BOOK   Y 

Measurement  of  the  Circle  .        .        .        .        o        .        .  220 

Additional  Exercises »        .  243 

5 


INDEX   OF   MATHEMATICAL   TERMS 

[The  references  are  to  articles] 


Abbreviations,  list  of,  29 
Acute  angle,  17 
Adjacent  angles,  15 
Alternate  exterior  angles,  113 
Alternate  interior  angles,  113 
Alternation,  in  proportion,  416 
Altitude,  of  parallelogram,  561 

of  trapezoid,  561 

of  triangle,  194 
Analysis,  246 
Angle,  13 

acute,  17 

at  center  of  a  regular  polygon, 

degree  of,  347 

inscribed,  354 

oblique,  17 

obtuse,  17 

right,  16 

sides  of,  13 

vertex  of,  13 
Angles,  adjacent,  15 

alternate  exterior,  113 

alternate  interior,  113 

complementary,  (yQ 

corresponding,  113 

exterior,  113 

homologous,  544 

interior,  113 

opposite,  75 

supplementary,  65 

vertical,  75 
Antecedents,  in  proportion,  404 
Apothem,  716 
Arc,  21 

degree  of,  347 
Area,  561 
Axiom,  27 
Axioms,  list  of,  27 


716 


Base,  of  isosceles  triangle,  50 
of  parallelogram,  561 
of  polygon,  561 
of  trapezoid,  561 

Center,  of  circle,  19 

of  regular  polygon,  716 

of  similitude,  556 
Chord,  247 
Circle,  19 

angle  inscribed  in,  354 

center  of,  19 

circumference  of,  19 

diameter  of,  20 

inscribed  in  polygon,  247 

radius  of,  20 

sector  of,  343 

segment  of,  354 

tangent  to,  306 
Circumscribed  circle,  247 

polygon,  247 
Commensurable  quantities,  342 
Complementary  angles,  66 
Composition  and  division,  430 
Composition,  in  proportion,  421 
Conclusion,  22 
Concurrent  lines,  601 
Consequents,  in  proportion,  404 
Constant,  340 
Continued  proportion,  442 
Converse,  72 
Corollary,  25 

Corresponding  angles,  113 
Curved  line,  7 

Decagon,  152 
Degree,  of  angle,  347 
of  arc,  347 


6 


INDEX   OF  MATHEMATICAL    TERMS 


Determination  of  straight  line,  246 

Diagonal,  152 

Diameter,  20 

Direct  tangent,  656 

Distance;  from  point  to  line,  223 

from  point  to  point,  223 
Division,  external,  506 

in  proportion,  427 

internal,  506 
Dodecagon,  152 

Equiangular  polygon,  152 

triangle,  18 
Equilateral  polygon,  152 

triangle,  18 
Equivalent  polygons,  561 
Exterior  angles,  113 
External  division,  506 
Extreme  and  mean  ratio,  551 
Extremes,  in  proportion,  404 

Fourth  proportional,  404 

Geometrical  figures,  9 
Geometry,  11 
plane,  12 

Harmonical  division,  509 

pencil,  512 
Hexagon,  152 
Homologous  angles,  544 

sides,  544 
Hypotenuse,  43 
Hypothesis,  22 

Incommensurable  quantities,  342 
Indirect  proof,  39 
Inscribed  angle,  354 

circle,  395 
Interior  angles,  113 
Inversion  in  proportion,  419 
Isosceles  triangle,  18 
Isosceles  triangle,  base  of,  50 

vertex  of,  50 


Left  side,  of  angle,  130 
Legs,  of  right  triangle,  43 
Limit,  340 
Line,  4 

curved,  7 

straight,  6 
Lines,  parallel,  107 

perpendicular,  16 
Locus,  233 

Material  body,  1 

Mean  proportional,  404 

Means,  of  a  proportion,  404 

Median  of  a  triangle,  173 

Minutes,  of  arc,  347 

Mutually  equiangular  triangles,  137 

Oblique  angle,  17 
Obtuse  angle,  17 
Octagon,  152 

Parallel  lines,  107 
Parallelogram,  194 

altitude  of,  561 

bases  of,  561 
Pentagon,  152 
Pentedecagon,  152 
Perimeter  of  polygon,  152 
Perpendicular  lines,  16 
Plane  surface,  8 

angle,  13 

figure,  10 

geometry,  12 
Point,  5 
Polar,  521 
Pole,  521 
Polygon,  152 

circumscribed,  395 

diagonal  of,  152 

inscribed,  247 

perimeter  of,  152 

regular,  152 
Polygons,  similar,  477 


8 


INDEX   OF  MATHEMATICAL    TERMS 


Postulate,  28 
Problem,  23 
Projection,  657 
Proportion,  404 

antecedents  of,  404 

consequents  of,  404 

couplets,  404 

extremes  of,  404 

means  of,  404 
Proportional,  fourth,  404 

mean,  404 

third,  404 
Proposition,  24 

Quadrarft,  346 

Quadrilateral,  152 

Quantities,  commensurable,  342 

constant,  340 

incommensurable,  342 

variable,  340 

Radius,  20 
Ratio,  340 

extreme  and  mean,  551 
Rays,  of  pencil,  512 
Rectangle,  194 
Regular  polygon,  707 

angle  at  center  of,  7 16 

apothem  of,  716 

center  of,  716 
Rhomboid,  194 
Rhombus,  194 
Right  angle,  16 
Right-angled  triangle,  18 
Right  side  of  angle,  130 

Scalene  triangle,  18 
Scholium,  26 
Secant,  262 
Seconds,  of  arc,  347 
Sector,  343 
Segment,  354 
angle  inscribed  in,  354 


Sides  of  angle,  13 
of  triangle,  18 

Similar  arcs,  787 
polygons,  477 
sectors,  787 
segments,  787 

Similitude,  center  of,  556 

Solid,  geometrical,  2 

Square,  194 

Straight  line,  6 
determined,  246 
divided  externally,  506 
divided  harmonically,  509 
divided  internally,  506 

Supplementary  angles,  65 

Surface,  3 

Symbols,  list  of,  29 

Tangent  circles  and  lines, 

306 
Tangent,  direct,  556 

transverse,  556 
Theorem,  22 
Third  proportional,  404 
Transversal,  113 
Transverse  tangent,  556 
Trapezium,  194 
Trapezoid,  194 

altitude  of,  561 

bases  of,  561 
Triangle,  18 

altitude  of,  194 

equiangular,  18 

equilateral,  18 

isosceles,  18 

median  of,  173 

right-angled,  18 

scalene,  18 

Variable,  340 
Vertex  of  angle,  13 
Vertical  angles,  75 
in  isosceles  triangle,  50 


PLANE    GEOMETRY 

DEFINITIONS 

1.  Every  material  body  occupies  a  limited  portion  of  space. 
If  we  conceive  the  body  to  be  removed,  the  space  that  is  left, 
which  is  identical  in  form  and  magnitude  with  the  body,  is  a 
geometrical  solid. 

2.  A  geometrical  solid,  then,  is  a  limited  portion  of  space.  It 
has  three  dimensions :  length,  breadth,  and  thickness. 

3.  The  boundaries  of  a  solid  are  surfaces.  A  surface  has 
but  two  dimensions :  length  and  breadth. 

4.  The  boundaries  of  a  surface  are  lines.  A  line  has  length 
only. 

5.  The  ends  of  a  line  are  points.  A  point  has  position,  but 
no  magnitude. 

6.  A  straight  line  is  one  that  does  not  change  its  direction  at 
any  point. 

7.  A  curved  line  changes  its  direction  at  every  point. 

8.  A  plane  surface  is  a  surface,  such  that  a  straight  line 
joining  any  two  of  its  points  will  lie  wholly  in  the  surface.    • 

9.  Any  combination  of  points,  lines,  surfaces,  or  solids,  is  a 
geometrical  figure. 

10.    A  figure  formed  by  points  and  lines  in  a  plane   is  a 
plane  figure. 


10 


PLANE   GEOMETRY 


11.  Geometry  is  the  science  that  treats  of  the  properties, 
the  construction,  and  the  measurement  of  geometrical  figures. 

12.  Plane  Geometry  treats  of  plane  figures. 


13.  A  plane  angle  is  the  amount  of  divergence  of  two  lines 
that  meet.  The  lines  are  the  sides  of  the  angle,  and  their 
point  of  meeting  is  the  vertex. 

One  way  to  indicate  an  angle  is  by  the  use  of  three  letters. 
Thus,  the  angle  in  the  accompanying  fig- 
ure is  read  angle  ABC  or  angle  CBA,  the 
letter  at  the  vertex  being  in  the  middle. 

If  there  is  only  one  angle  at  the  ver- 
tex B,  it  may  be  read  angle  B. 

Another  way  is  to  place  a  small  figure  or  letter  within  the 
angle  near  the  vertex.     The  above  angle  may  be  read  angle  S. 

The  size  of  an  angle  in  no  way  depends  upon  the  length  of 
its  sides,  and  is  not  altered  by  either  increasing  or  diminishing 
their  length. 


14.  Two  angles  are  equal  if  they  can 
be  made  to  coincide.  Thus,  angles  ABC  and 
DEF  are  equal,  whatever  may  be  the  length 
of  each  side,  if  angle  ABC  can  be  placed 
upon  angle  DEF  so  that  the  vertex  B  shall 
fall  upon  vertex  E,BC  fall  upon  EF,  and  BA 
fall  upon  ED. 

[It  should  be  noticed  that  angle  ABC  can  be 
made  to  coincide  with  angle  DEF  in  another  way, 
i.e.  ABC  may  be  turned  over  and  then  placed 
upon  DEF,  EG  falling  upon  ED,  and  BA  upon 
EF.-\ 

15.  Two  angles  that  have  the  same  ver- 
tex and  a  common  side  are  adjo/cent  angles. 

Angles  1  and  2  are  adjacent  angles. 


DEFINITIONS 


11 


16.   If   a   straight   line   meets  another  straight  line  so  as 
to  make  the  adjacent  angles  that  they 
form  equal  to  each   other,  the   angles 
formed  are  right  angles.     Angles  ABC 
and   ABD   are    right    angles.      In  this 

case  each  line  is  perpendicular  to  the         

other. 


17.  An  angle  that  is  less  than  a  right 
angle  is  acute,  and  one  that  is  greater 
than  a  right  angle  is  obtuse. 

An  angle  that  is  not  a  right  angle  is 
oblique. 


18.  A  triangle  is  a  portion  of  a  plane 
bounded  by  three  straight  lines.  The 
lines  are  called  the  sides  of  the  tri- 
angle, and  their  angles  the  angles  of 
the  triangle. 

An  equilateral  triangle  has  three  equal 
sides. 

An  isosceles  triangle  has  two  equal  sides. 

A  scalene  triangle  has  no  two  sides  equal. 

An  equiangular  triangle  has  three  equal  angles. 

A  right-angled  triangle  contains  one  right  angle, 


19.  A  circle  is  a  portion  of  a  plane  bounded 
by  a  curved  line,  all  points  of  which  are 
equally  distant  from  a  point  within,  called 
the  center.  The  bounding  line  is  called  the 
circumference. 

20.  The  distance  from  the  center  to  any 
point  on  the  circumference  is  a  radius. 


21.    Any  portion  of  a  circumference  is  an  arc. 


12  PLANE   GEOMETRY 

22.  A  theorem  is  a  truth  requiring  demonstration.  The 
statement  of  a  theorem  consists  of  two  parts,  the  hypothesis 
and  the  conclusion.  The  hypothesis  is  that  part  which  is 
assumed  to  be  true;  the  conclusion  is  that  which  is  to  be 
proved. 

23.  A  problem  proposes  to  effect  some  geometrical  construc- 
tion, such  as  to  draw  some  particular  line,  or  to  construct  some 
required  figure. 

24.  Theorems  and  problems  are  called  propositions. 

25.  A  corollary  is  a  truth  that  may  be  readily  deduced  from 
one  or  more  propositions, 

26.  A  scholium  is  a  remark  made  upon  one  or  more  proposi- 
tions relating  to  their  use,  connection,  limitation,  or  extension. 

27.  An  axiom  is  a  self-evident  truth. 


Axioms 

1.  Things  that  are  equal  to  the  same  thing  are  equal  to  each 
other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal. 

4.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 
.   5.    If  equals  are  divided  by  equals,  the  quotients  are  equal. 

6.  If  equals  are  added  to  unequals,  the  sums  are  unequal  in 
the  same  order. 

7.  If  equals  are  subtracted  from  unequals,  the  remainders 
are  unequal  in  the  same  order. 

8.  If  unequals  are  multiplied  by  positive  equals,  the  prod- 
ucts are  unequal  in  the  same  order. 


DEFINITIONS  13 

.    9.   If  unequals  are  divided  by  positive  equals,  the  quotients 
are  unequal  in  the  same  order. 

10.  If  unequals  are  added  to  unequals,  the  greater  to  the 
greater,  and  the  less  to  the  less,  the  sums  are  unequal  in  the 
same  order. 

11.  The  whole  is  greater  than  any  of  its  parts. 

12.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

13.  Only  one  straight  line  can  be  drawn  joining  two  points. 

[It  follows  from  this  axiom  that  two  straight  lines  can  intersect  in  only 
one  point.] 

14.  The  shortest  distance  from  one  point  to  another  is  meas- 
ured on  the  straight  line  joining  them. 

15.  Through  a  point  only  one  line  can  be  drawn  parallel  to 
another  line. 

16.  Magnitudes  that  can  be  made  to  coincide  with  each 
other  are  equal. 

[This  axiom  affords  the  ultimate  test  of  the  equality  of  geometrical 
magnitudes.  It  implies  that  a  figure  can  be  taken  from  its  position,  with- 
out change  of  form  or  size,  and  placed  upon  another  figure  for  the  purpose 
of  comparison.] 

Of  the  foregoing,  the  first  twelve  axioms  are  general  in  their 
nature,  and  the  student  has  probably  met  with  them  before  in 
his  study  of  algebra.  The  last  four  are  strictly  geometrical 
axioms. 

28.  A  postulate  is  a  self-evident  problem. 

Postulates 

1.  A  straight  line  can  be  drawn  joining  two  points. 

2.  A  straight  line  can  be  prolonged  to  any  length. 

3.  If  two  lines  are  unequal,  the  length  of  the  smaller  can  be 
laid  off  on  the  larger. 

4.  A  circumference  can  be  described  with  any  point  as  a 
center,  and  with  a  radius  of  any  length. 


14 


PLANE   GEOMETRY 


29.                     SYMBOLS  AND 

Z  Angle. 

A  Angles. 

R.A.  Right  angle. 

R.A.'s.  Right  angles. 

A  Triangle. 

k.  Triangles. 

O  Circle. 

©  Circles. 

±  Perpendicular. 

Js  Perpendiculars. 

11  Parallel. 

lis  Parallels. 

ABBREVIATIONS 

.-.  Therefore. 

=  Equals  or  equal. 

>  Is  (or  are)  greater  than. 

<  Is  {or  are)  less  than. 

~  Is  (or  are)  measured  by. 

Prop.  Proposition. 

Cor.  Corollary. 

Schol.  Scholium. 

Q.E.D.  Quod  erat  demonstrandum, 
which  was  to  be  proved. 

Q.E.r.  Quod  erat  faciendum,  which 
was  to  be  done. 


BOOK  I 


Proposition  I.     Theorem 


30.  If  two  triangles  have  two  sides  and  the  included 
angle  of  one  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other,  the  triangles  are  equal  in 
all  respects. 


Let  the  A  ABC  and  DBF  have  AB  =  DE^  BC  =  EF^  and 
Z.B  =  /.E. 

To  Prove  the  A  ABC  and  DEF  equal  in  all  respects. 

Proof.  Place  the  A  ABC  upon  the  A  DEF  so  that  Z  B  shall 
coincide  with  its  equal  Z  E,  BA  falling  upon  ED,  and  BC  upon 
EF. 

Since,  by  hypothesis,  BA  =  ED,  the  vertex  A  will  fall  upon 
the  vertex  D. 

Since,  by  hypothesis,  BC  =  EF,  the  vertex  c  will  fall  upon 
the  vertex  F. 

Since,  by  Axiom  13,  only  one  straight  line  can  be  drawn 
joining  two  points,  ^(7  will  coincide  with  DF.  ,\  the  A  coin- 
cide throughout  and  are  equal  in  all  respects.  q.e.d. 

15 


16 


PLANE   GEOMETRY 


31.  Scholium.  By  showing  that  the  A  coincide,  we  have 
not  only  proved  that  they  are  equal  in  area,  but  also  that 
ZA  =  Z.D,  /.C=/.F,  and  AC  =  DF. 

It  should  be  noticed  that  the  sides  AC  and  DF,  which  have 
been  proved  equal,  lie  opposite  respectively  to  the  equal  angles 
B  and  E. 

Also,  that  the  equal  angles  A  and  D  lie  opposite  respec- 
tively to  the  equal  sides  BC  and  EF,  and  that  the  equal 
angles  C  and  F  lie  opposite  respectively  to  the  equal  sides 
AB  and  DE. 

Principle.  In  triangles  that  have  been  proved  equal  in  all 
respects,  equal  sides  lie  opposite  equal  angles,  and  equal  angles 
lie  opposite  equal  sides. 

D  E 


32.   Exercise.    Prove  Prop.  I. ,  using 
this  pair  of  triangles. 


33.  Exercise.     In  the  triangle  ABC,  AB  = 
AC,  and  AD  bisects  the  angle  BAG. 
Prove  that  AD  also  bisects  BG. 

Suggestion.  Show  by  §  30  that  the  A  ABD 
and  ADC  are  equal  in  all  respects.  Then,  by 
the  principle  of  §  31,  BD  =  DG. 


34.    Exercise.     ABC  is  a  triangle  having 
AB  =  BC.     BE  is  laid  off  equal  to  BD. 

AD  =  CE. 


Prove 


ion.     Show  that 

AABD  =  AEBC. 


BOOK  I  17 


Proposition  II.     Theorem 


35.  If  two  triangles  have  two  angles  and  the  included 
side  of  one  equal  respectively  to  two  angles  and  the 
included  side  of  the  other,  the  triangles  are  equal  in 
all  respects. 


Let  the  A  ABC  and  DBF  have  Z.A=/.D,  /.c  =  Z.F,  and 

AC  =  DF. 

To  Prove  the  /^  ABC  and  DEF  equal  in  all  respects. 

Proof.  Place  the  A  ABC  upon  the  A  DEF,  so  that  Z  A  shall 
coincide  with  its  equal  Z.D,  AB  falling  upon  DE,  and  AC 
falling  upon  DF. 

Since,  by  hypothesis,  AC  =  DF,  the  vertex  c  will  fall  upon 
vertex  F. 

Since,  by  hypothesis,  Z  C  =  Zf,  the  side  CB  will  fall  upon 
FE,  and  the  vertex  B  will  be  on  FE  or  its  prolongation. 

Since  AB  falls  upon  DE,  the  vertex  B  will  be  upon  DE  or 
its  prolongation. 

The  vertex  B,  being  at  the  same  time  on  DE  and  FE,  must 
be  at  their  point  of  intersection ;  and  since  two  straight  lines 
have  only  one  point  of  intersection  (Axiom  13),  the  vertex  B 
must  fall  at  E. 

.'.  the  A  ABC  and  DEF  coincide  throughout,  and  are  equal 
in  all  respects.  q.e.d. 

SANDERS^    GEOM.  —  2 


18 


PLANE   GEOMETRY 


AwD 


36.    Exercise.   Prove  Prop.  II.,  using  this 
pair  of  triangles. 


37.  Exercise.  In  the  A  ^50,  BD  bisects 
/.ABC  and  is  perpendicular  to  AC. 

Prove  that  BD  bisects  AC  and  that  AB 
=  BC. 


38.    Exercise.    ABC  is  a  A  having  Z  BAG  ^ 
=  BCA.     AD  bisects  ZBAC  and  CU  bisects 
ZBCA. 

Prove  AD  =  CE. 

Suggestion.  Prove  ^  ^DC  and  AEC  equal 
in  all  respects  by  §  35.  Then  by  the  Principle 
of  §  31,  AD  =  EC.  C 


39.  The  next  proposition  is  an  example  of  what  is  called  the 
indirect  proof. 

The  reasoning  is  based  on  the  following  Principle :  If  the 
direct  consequences  of  a  certain  supposition  are  false,  the  supposi- 
tion itself  is  false. 

To  prove  a  theorem  by  this  plan,  the  following  steps  are 
necessary : 

1.  The  theorem  is  supposed  to  be  untrue. 

2.  The  consequences  of  this  supposition  are  shovru  to  be 
false. 

3.  Then,  by  the  above  Principle,  the  supposition  (that  the 
theorem  is  untrue)  is  false. 

4.  The  theorem  is  therefore  true. 


BOOK  I 


19 


Proposition  III.     Theorem 

40.   At    a  given   point    in   a   line    only   one   perpen- 
dicular can  he  erected  to  that  line. 


Let  CD  be  ±  to  ab  at  the  point  D. 

To  Prove  CD  is  the  only  _L  that  can  be  erected  to  AB  at  D. 

Proof.  Suppose  a  second  ±,  as  DE,  could  be  erected  to  AB 
at  D. 

By  hypothesis  and  §  16,         Z.  CD  A  =  Z.  CDB. 

By  supposition  and  §  16,       /.  EDA  =  Z  EDB. 

But         Z  EDA  >  Z  CD  A,  and  Z  EDB  <  Z  CDB. 

.-,  Z  EDA  cannot  equal  Z  EDB,  and  DE  cannot  be 
±   to   AB. 

The  supposition  that  a  second  ±  could  be  erected  to  AB  at  D 
is  therefore  false,  and  only  one  _L  can  be  erected  to  AB  at  that 
point.  Q.E.D. 

Note.  The  points  and  lines  of  the  above  figure,  and  of  all  figures 
given  in  the  first  five  books  of  this  geometry,  are  understood  to  be  in 
the  same  plane.  The  term  "line"  is  used  in  this  work  for  "straight 
line." 


20  PLANE    GEOMETRY 

41.    CoBOLLARY.     All  right  angles  are  equal. 
A  D  G 


Let  /.ABC  and  Z  def  be  2  R.A/s. 
To  Prove  /.ABG  =  A  DEF. 

Proof.  Suppose  them  to  be  unequal  and  that  /.ABC,  when 
superimposed  upon  /  I)EF,  takes  the  position  GEF. 

Then  at  E  there  would  be  two  perpendiculars  to  EF,  which 
contradicts  §  40. 

Therefore  the  supposition  that  the  right  angles  ABC  and 
DEF  are  unequal  is  false,  and  they  are  equal.  q.e.d. 

42.  Scholium.  The  right  angle  is  the  unit  of  measure  for 
angles.  An  angle  is  generally  expressed  in  terms  of  the  right 
angle.     Thus,  Z  ^  =  |  E.A.,  or  /  B  =  1\  R.A.,  etc. 

43.  Definitions.  In  a  right-angled 
triangle  the  side  opposite  the  right 
angle  is  called  the  hypotenuse. 

The  other  two  sides  are  the  legs  of 
the  triangle. 

44.  Exercise.  If  two  B.A.  ^  have  the  legs  of  one  equal  respectively 
to  the  legs  of  the  other,  the  A  are  equal  in  all  respects. 

45.  Exercise.  A  is  40  miles  west  of  B.  C  is  30  miles  north  of  A^ 
and  D  is  30  miles  south  of  A.  From  C  to  5  is  50  miles.  How  far  is  it 
from  DtoB? 

46.  Exercise.  A  is  m  yards  north  of  B.  G  is  n  yards  west  of  A, 
and  Disn  yards  east  of  B.  Prove  that  the  distance  from  ^  to  O  is  the 
same  as  the  distance  from  A  to  D. 


BOOK  I  21 

Proposition  IV.     Theorem 

47.    If  a  perpendicular   is  drawn   to   a   line  at   its 
middle  point, 

I.    Any  point  on  the  perpendicular  is  equally  dis- 
tant from  the  extremities  of  the  line. 

II.    Any    point    without    the    perpendicular    is    un- 
equally distant  from  the  extremities  of  the  line. 


I.   Let  CD  be  ±  to  AB  at  its  middle  point  D,  and  P  be  any 
point  on  CD. 

To  Prove  P  equally  distant  from  A  and  B. 

Draw  PA  and  PB. 

[It  is  required  to  prove   PA  =  PB,    for  PA  and   PB  measure  the 
distance  from  P  to  A  and  B  respectively.] 

Proof.     The  A  PAD  and  PBD  have 

AD  =  DB  (Hypothesis), 

Z  1  =  Z  2  (Right  Angles), 

PD  =  PD  (Common). 
The  A  are  equal  in  all  respects  by  §  30. 
,-.  PA  =  PB,  and  P  is  equally  distant  from  A  and  B.       q.e.d. 


22  '  PLANE    GEOMETRY 

C 


II.  Let  <JB  be  ±  to  AB  at  its  middle  point  D,  and  P  be  any 
point  without  CD. 

To  Prove  P  unequally  distant  from  A  and  B. 

Draw  P^  and  PP. 

[It  is  required  to  prove  PA  and  FB  unequal.  ] 

Proof.     One  of  these  lines,  as  PA,  will  intersect  the  perpen- 
dicular GB  in  some  point,  as  E. 
Draw  EB. 

PB<PE^  EB.  Axiom  14. 

EB  =  EA.  By  Case  I. 

Substitute  EA  for  ZP. 

PB  <PE  -\-  EA. 

PB  <  PA 

(PE  and  ^^  make  up  PA). 

Since  PP  and  PA  are  unequal,  P  is  unequally  distant  from 
A  and  P.  Q.E.D. 

48.  Corollary  I.  A  perpendicular  erected  to  a  line  at  its 
middle  point  contains  all  points  thai  are  equally  distant  from  the 
extremities  of  the  line. 

For,  by  §  47,  all  points  on  the  perpendicular  are  equally 
distant  from  the  extremities  of  the  line,  and  all  points  without 
the  perpendicular  are  unequally  distant  from  the  extremities 
of  the  line.  Therefore  all  points  that  are  equally  distant  from 
the  extremities  of  the  line  must  be  on  the  perpendicular. 


BOOK  I 


23 


49.  Corollary  II.  If  a  tine  has  two  of  its  points  each 
equally  distant  from  the  extremities  of  another  line,  the  first  line 
is  perpertdicular  to  the  second  at  its  middle  point. 

Let  AB  have  two  of  its  points  m  and 
n  each  equally  distant  from  the  extrem- 
ities of  CD. 

To  Prove  AB  ±  to  CD  at  its  middle 
point.  C  "^^^ 

Proof.  Suppose  a  line  were  drawn 
±  to  CD  at  its  middle  point. 

By  §  48  both  m  and  n  must  be  on 
this  perpendicular. 

By  hypothesis  both  m  and  n  are  on  AB. 

So  the  perpendicular  and  AB  both  pass  through  m  and  n. 

By  Axiom  13  only  one  straight  line  can  pass  through  two 
given  points. 

.',  AB  must  coincide  with  the  perpendicular 
to  CD  at  its  middle  point.  q.e.d. 


50.  Definitions.  In  an  isosceles  triangle 
the  angle  formed  by  the  two  equal  sides  is 
called  the  vertical  angle.  The  side  opposite 
this  angle  is  usually  called  the  base  of  the 
triangle. 


51.  Exercise.  If  a  perpendicular  is  erected  to  the  base  of  an  isosceles 
A  at  its  middle  point,  it  passes  through  the  vertex  of  the  vertical  angle. 

Suggestion.     Use  §  48.  • 

52.  Exercise.  If  a  line  is  drawn  from  the  vertex  of  the  vertical 
angle  of  an  isosceles  A  to  the  middle  point  of  the  base,  it  is  perpendicular 
to  the  base. 

Suggestion.     Use  §  49. 


24 


PLANE   GEOMETRY 


Proposition  V.     Problem 

53.   To   erect   a  perpendicular  to  a  line  at  a  given 
point  on  that  line. 


■B 


Let  AB  be  the  given  line,  and  C  the  given  point  on  the  line. 

Required  to  erect  a  perpendicular  to  AB  at  C. 

Lay  off  CD  =  CE. 

With  D  and  E  as  centers,  and  with  a  radius  greater  than 
DC  (one  half  of  DE),  describe  two  arcs  intersecting  at  F. 

Join  F  and  C. 

FC  is  the  required  perpendicular.  For,  F  and  C  are  each 
equally  distant  from  D  and  E  (construction).  .".  by  §  49,  FC 
is  perpendicular  to  DE  or  AB.  q.e.f. 

54.  Exercise.  To  construct  a  R.A.  A,  having  given  the  two  sides 
about  the  R.A. 

Let  m  and  n  be  the  two  given  sides. 

Required  to  construct  a  R.A. 
A,  having  m  and  n  as  sides  about          ~^—~^~ 
the  R.A.  H 

Lay  off  the  indefinite  line  AB. 

At  any  point  of  it  as  C  erect 
CD  ±  AB^  and  make  CD  equal 
in  length  to  m. 

Lay  off  CE  equal  to  n. 

Draw  DE. 

A  CDE  is  the  required  A  because  it  fulfills  all  the  required  conditions  ; 
i.e.  it  is  right  angled  at  C,  and  the  sides  about  0  are  equal  respectively  to 
m  and  n.  q.e.f 


BOOK  I 


25 


Proposition  VI.     Prob.lem 
55.  To  bisect  a  given  line. 


/  \ 


M 


H^ 


Let  AB  be  the  given  line. 

Required  to  bisect  it. 

With  A  and  B  as  centers,  and  with  any  radius  greater  than 
one  half  oi  AB,  describe  arcs  intersecting  at  c  and  D. 

Draw  CD. 

Then  will  CD  bisect  AB. 

For,  the  points  C  and  D  are  each  equally  distant  from  the 
extremities  of  AB  (construction).     .-.  CD  bisects  AB  (§  49). 

Q.E.F. 

56.  Exercise.     Divide  a  given  line  into  quarters. 

57.  Exercise.  If  the  radius  used  for  describing  the  two  arcs  that 
intersect  at  C  in  the  figure  of  Prop.  VI  is  greater  tlian  the  radius  used 
for  describing  the  two  arcs  that  intersect  at  Z>,  will  CD  bisect  AB  ? 

58.  Exercise.     When  will  the  lines  AB  and  CD  bisect  each  other  ? 

59.  Exercise.  In  a  given  line  find  a  point  that  is  equally  distant 
from  two  given  points.     When  is  this  problem  impossible  ? 


26 


PLANE   GEOMETRY 


Proposition  YII.     Theorem 

60.  The  sum   of  the  adjacent  angles  formed  hy  one 
line  meeting  another,  is  two  right  angles. 


B 


Let  AB  meet  CD  at  5. 

To  Prove  /.ABC -\-/.  abb  =  2  R.A.'s. 

Proof.     Erect  BE  perpendicular  to  CD  at  B.     (§  53.) 
By  construction  /.EBC  and  Z.EBD  are  R.  A.'s. 

/.ABC=1  R.A.  -\-/EBA. 

Z  ABD  =  1  R.A.  —  Z  EBA. 

Adding  (1)  and  (2),  Aabc-\-Z  abd  =  2  R.A's. 


(1) 
(2) 

Q.E.D. 


61.  Corollary  I.  If  one  of  two  adjacent  angles  formed  by 
one  line  meeting  another  is  a  right  angle,  the  other  is  also  a  right 
angle. 

62.  Corollary  II.  If  two  straight  lines  intersect  each  other, 
and  one  of  the  angles  formed  is  a  right  angle,  the  other  three 
angles  are  also  right  angles. 

63.  Corollary  III.  The  sum  of  all  the  angles  formed  at  a 
point  in  a  line,  and  on  the  same  side  of  the  line,  is  two  right 
angles.  p 

Suggestion.    Show  that  the  sum  of  all  D». 

the  angles  at  C  equals 

/.FCA  +  Z.FCB 

A ^¥f^ B 


or 


ZGCA  +  ZGCB,  etc. 


BOOK  I 


27 


64.  Corollary  IV.  TJie  sum  of 
all  the  angles  formed  about  a  j)oint 
is  four  right  angles. 

Suggestion.  Prolong  one  of  the  lines, 
as  OE,  to  a.  Then  apply  §  63  to  the 
angles  on  each  side  of  GE. 

65.  Definition.  If  two  angles 
are  together  equal  to  two  right 
angles,  they  are  called  supplemen- 
tary angles.  Each  angle  is  the  sup- 
plement of  the  other. 

Adjacent  angles  formed  by  one  line  meeting  another  are 
supplementary  adjacent  angles. 

66.  Definition.  If  two  angles  are  together  equal  to  one 
right  angle,  they  are  called  complementary  angles.  Each  angle 
is  the  complement  of  the  other. 

67.  Exercise.  Find  the  supplement  and  also  the  complement  of  each 
of  the  following  angles  :  |  R.A.,  ^  R.  A,,  |  R.A. 

Find  the  value  of  each  of  two  supplementary  angles,  if  one  is  five  times 
the  other. 


68.    Exercise. 
complement. 


Given  an  angle,  construct  its  supplement  and  also  its 


69.  Exercise.  Prove  that  the  bisectors 
of  two  supplementary  adjacent  angles  are 
perpendicular  to  each  other. 

70.  Exercise.  Through  the  vertex  of  a 
right  angle  a  line  is  drawn  outside  of  the 
angle.  What  is  the  sum  of  the  two  acute 
angles  formed  ?     [Zl+Z2  =  ?] 

71.  Exercise.    Find  the  supplement  of  the  complement  of  |  R.A., 
also  the  complement  of  the  supplement  of  If  R.  A. 


28  PLANE   GEOMETRY 

72.  Definition.  One  proposition  is  the  converse  of  another, 
when  the  hypothesis  and  conclusion  of  one  are  respectively 
the  conclusion  and  hypothesis  of  the  other. 

The  converse  of  a  proposition  is  not  necessarily  true. 

We  shall  prove  later  (see  §  85)  that  "if  the  sides  of  one 
triangle  are  equal  respectively  to  the  sides  of  another,  the 
angles  of  the  first  triangle  are  equal  respectively  to  those  of 
the  second." 

Show,  by  drawing  triangles,  that  the  converse  of  this  propo- 
sition, i.e.  "  if  the  angles  of  one  triangle  are  equal  respectively 
to  the  angles  of  another,  the  sides  of  the  first  triangle  are  equal 
respectively  to  those  of  the  second,"  is  not  necessarily  true. 

Proposition  VIII.   Theorem  (Converse  of  Prop.  VII.) 

73.  If  the  sum  of  two  adjacent  angles  is  two  right 
angles,  their  exterior  sides  form  a  straight  line. 

,0 


Let  Z  CD  A  +  Z  CDB  =  2  K.A.'s. 

To  Prove  AD  and  DB  form  a  straight  line. 

Proof.  Suppose  DB  is  not  the  prolongation  of  AD,  and  that 
some  other  line,  as  DE,  is. 

By  §  60  Z  CDA  +  Z  CDE  would  equal  2  R.A.'s. 

By  hypothesis  Z  CDA  +  Z  CDB  =  2  R.A.'s. 

By  Axiom  1.,  Z  CDA  -\-  Z  CDE  would  equal  Z  CDA  +  Z  CDB. 

Whence  Z  CDE  would  equal  Z  CDB. 

This  contradicts  Axiom  11. 

Therefore  the  supposition  that  DB  is  not  the  prolongation  of 
AD  is  false,  and  AD  and  DB  form  a  straight  line.  q.e.d. 


BOOK  1 


29 


74.  Exercise.  ABO 
and  DEF  are  R.A.  A 
equal  in  all  respects, 
right  angled  at  B  and  D.  (^ 
Place  A  ABC  in  the 
position  of  A  GED. 

Prove  that  GD  and  DF  form  a  straight  line. 

75.  Definition.  If  two  lines  inter- 
sect each  other,  the  opposite  angles 
formed  are  called  vertical  angles.  Z.  1 
and  Z  3  are  vertical  angles,  as  are  also 
Z2  and  Z4. 

76.  Exercise.  The  bisectors  of  two 
opposite  angles  form  a  straight  line. 

Let  FE,  HE,  GE,  and  JE  be  the  bi- 
sectors of  A  AEG,  CEB,  BED,  and  BE  A 
respectively. 

To  Prove  that  FE  and  EG  form  a 
straight  line,  and  HE  and  EJ  form  a 
straight  line. 

Suggestion.     Use  §§  69  and  73. 

Proposition  IX.     Theorem 

77.  If  two  straight  lines  intersect,  the  opposite  or  ver- 
tical angles  are  equal. 

Let  AB  and  CD  intersect. 

To  Prove  Z 1  =  Z  3 
and  Z  2  =  Z  4. 

Proof. 
Zl+Z2=2E.A.'s. 

(authority  ?) 

Z2-f-Z3  =  2R.A.'s.      (?) 

Z1+Z2  =  Z2+Z3.   (?) 

Zlz=Z3.  (?) 

In  the  same  manner  prove  Z  2  =  Z  4.  q.e.d. 


30 


PLANE   GEOMETRY 


78.  Exercise.  One  angle  formed  by  two  intersecting  lines  is  |  R.A. 
Find  the  other  three. 

79.  Exercise.     The  bisector  of  an  angle  bisects  its  vertical  angle. 

80.  Exercise.  Two  lines  intersect,  making  the  sum  of  one  pair  of 
vertical  angles  equal  to  five  times  the  sum  of  the  other  pair  of  vertical 
angles.    Find  the  values  of  the  four  angles. 


Proposition  X.     Theorem 

81.  In  an  isosceles  triangle,   the  angles  opposite  the 
equal  sides  are  equal. 

Let  ABC  be  an  isosceles  A,  having 
AB  =BC. 

To  Prove  /.A  =  Z.C. 

Proof.     Draw  BD  bisecting  AC.     (§  55.) 

B  and  D  are  each  equally  distant  from  A 
and  C;  .-.  Zl  and  Z2  are  E.A.'s.     (?)  "  d 

Show  that  A  ABD  and  BBC  are  equal  in  all  respects. 

Whence  /-A  =  /.c.  q.e.d. 

82.  Corollary.     An  equilateral  triangle  is  equiangular. 


83.  Exercise.  ABC  is  an  isosceles  triangle. 
D  is  the  middle  point  of  the  base  AC  E  and  F 
are  the  middle  points  of  the  equal  sides  ^5  and  ^C. 

Prove  DE  =  DF. 


84.    Exercise.     ABC  is  an  isosceles  triangle 
having  AB=BC. 

BD  and  BE  are  drawn  making  Zl  =  Z2. 
Prove  Z3  =  Z4. 


BOOK  I 


81 


Proposition  XI.     Theorem 

85.  If  two  triangles  havo  three  sides  of  the  one  equal 
respectively  to  three  sides  of  the  other,  the  triangles  are 
equal  in  all  respects. 


Let  ABC  and  DEF  be  two  A,  having  AB  =  DE^  BC  =  EF,  and 
AC  =  DF. 

To  Prove  A  ABC  and  DEF  equal  in  all  respects. 

Proof.  Place  A  ABC  so  that  AC  shall  coincide  with  DF,  A 
falling  on  D  and  C  on  F,  and  the  vertex  B  falling  at  G,  on  the 
opposite  side  of  the  base  from  the  vertex  E. 

Draw  EG. 

Prove  Zl  =  Z2andZ3  =  Z4. 

Adding,  Z1  +  Z3  =  Z2  +  Z4,  orZ  DEF  =  Zdgf. 

Prove  A  DEF  and  DGF  equal  in  all  respects. 

.*.  A  DEF  and  ABC  are  equal  in  all  respects.  q.e.d. 

86.  Exercise.     Construct  a  triangle  having  given  its  three  sides. 

87.  Exercise.    Construct  a  triangle  equal  to  a  given  triangle. 

88.  Exercise,  Construct  a  triangle  whose  sides  are  in  the  ratio  of 
3,  4,  and  5. 


32 


PLANE    GEOMETRY 


Propositiox  XII.     Problem 

89.  To  draw  a  perpendicular  to  a  line  from  a  point 
without. 


C^-- 


.-^-'D 


Let  AB  be  the  given  line  and  P  the  point  without. 
Required  to  draw  a  perpendicular  from  P  to  the  line  AB. 

Let  s  be  any  point  on  the  opposite  side  of  AB  from  P. 

With  P  as  a  center,  and  Ps  as  a  radius,  describe  an  arc 
intersecting  AB  ?X  C  and  D. 

With  c  as  a  center,  and  with  a  radius  greater  than  one  half 
of  CD,  describe  an  arc ;  with  i)  as  a  center,  and  with  the  same 
radius,  describe  an  arc  intersecting  the  first  arc  at  E. 

Draw  PE. 

Show  that  PE  is  perpendicular  to  CD.  q.e.f. 


90.   Exercise. 
from  the  point  G. 


Draw  a  perpendicular  to  AB 


91.  Exercise.  If  the  line  AB  (see  §  89)  were 
situated  at  the  bottom  of  this  page,  and  there  were  no 
room  below  it  for  the  points,  how  could  the  perpen- 
dicular be  drawn  ? 


•C 


BOOK  I 


33 


Proposition  XIII.     Theorem 

92.  From  a  point  without  a  line  only  one  perpendicu- 
lar can  be  draivn  to  the  line. 


\ 

\ 


21 E 


fV 

Let  CD  be  a  ±  from  C  to  AB. 

To  Prove  that  CD  is  the  only  J_  that  can  be  drawn  from  C 
to  AB. 

Proof.     Suppose  a  second  J_,  as  CE,  could  be  drawn. 
Prolong  CD  until  DF  =  CD,  and  draw  EF. 
Prove  A  CDE  and  FDE  equal  in  all  respects. 

Whence  Z1  =  Z2. 

But  Z 1  =  1  R.A.  by  supposition. 

Show  that  Z 1  +  Z  2  =  2  R. A.'s. 

If  the  sum  of  angles  1  and  2  is  two  E-.A.'s,  CE  and  EF  form, 
a  straight  line.     (§  73.) 

The  points  C  and  F  are  therefore  connected  by  two  straight 
lines  (CDF  and  CEF),  which  contradicts  (?). 

Therefore  the  supposition  that  a  second  _L  could  be  drawn 
from  C  to  the  line  AB  is  false,  and  only  one  _L  can  be  drawn. 

Q.E.D. 

93.    Exercise.     Show  that  a  triangle  cannot  have  two  right  angles. 

SANDERS'    GEOM.  3 


34  PLANE   GEOMETRY 

Proposition  XIV.     Problem 
94.  To  bisect  a  given  angle. 


Let  ABC  be  any  angle. 
Required  to  bisect  it. 

With  ^  as  a  center,  and  with  any  convenient  radius,  describe 
an  arc  intersecting  the  sides  of  the  angle  at  D  and  E. 

With  Z)  as  a  center,  and  with  a  radius  greater  than  one  half 
of  DEj  describe  an  arc ;  with  ^  as  a  center,  and  with  the  same 
radius,  describe  an  arc  intersecting  this  arc  at  F. 

Join  B  and  F. 

Then  will  BF  bisect  Z.ABG. 

Draw  FE  and  FD. 

Prove  A  BEF  and  BDF  equal  in  all  respects. 

Whence  Z 1  =  Z  2,  and  /.  ABC  i^  bisected.  q.e.f. 

95.  Exercise.  At  a  given  point  on  a  line  construct  an  angle  equal 
to  I  R.A. 

96.  Exercise.    Divide  a  given  angle  into  quarters. 

97.  Exercise.  At  a  given  point  on  a  line  construct  an  angle  equal  to 
\l  R.A.'s. 

98.  Exercise.  Prove  §  81  by  drawing  BD  (see  figure  of  §  81)  bisect- 
ing angle  ABC. 

99.  Exercise.  Construct  a  triangle  ABC,  making  the  side  AB  two 
inches  long,  Z  ^  =  1  R.  A.  and  Z  5  =  ^  R.  A. 


BOOK  I  35 

Proposition  XV.     Problem 

100.  At  a  point  on  a  line  to  construct  an  angle  equal 
to  a  given  angle. 


C       D 


Let  /.ABC  be  the  given  angle,  and  F  the  point  on  the 
line  BE. 

Required  to  construct  an  angle  at  F  on  the  line  BE  that  shall 
equal  A  ABC. 

With  5  as  a  center,  and  with  any  radius,  describe  the 
arc  MG. 

With  i^  as  a  center,  and  with  the  same  radius,  describe  the 
indefinite  arc  LK^  intersecting  BE  at  K. 

With  iT  as  a  center,  and  with  the  distance  MG  as  a  radius, 
describe  an  arc  intersecting  the  arc  LK  afc  H. 

Draw  HF. 

Then  will  Z  HFK  =/.ABC. 

Draw  MG  and  HE. 

Prove  A  MB  G  and  HFK  equal  in  all  respects. 

Whence  /.B=/.F.  q.e.p. 

101.  Exercise.  Construct  a  triangle  having  given  two  sides  and  the 
included  angle. 

102.  Exercise.  Construct  a  triangle  having  given  two  angles  and 
the  included  side. 

103.  Exercise.  Construct  an  angle  equal  to  the  sum  of  two  given 
angles. 

104.  Exercise.     Construct  an  angle  that  is  double  a  given  angle. 


36  PLANE   GEOMETRY 

105.  ExERcisK.  Construct  an  angle  equal  to  the  difference  between 
two  given  angles. 

106.  Exercise.  Draw  any  triangle.  Construct  an  angle  equal  to 
the  sum  of  the  angles  of  this  triangle. 

From  your  drawing  what  do  you  infer  the  sum  of  the  angles  to  be  ? 
See  §  138. 

107.  Definition.  Parallel  lines  are  lines  lying  in  the  same 
plane,  which  do  not  meet,  how  far  soever  they  may  be 
prolonged. 

Proposition  XVI.     Theorem 

108.  If  two  lines  are  parallel  to  a  third  line,  they  are 

parallel  to  each  other. 


A 

C 

E 

Let  AB  and  CD  be  II  to  EF. 

To  Prove  AB  and  CD  II  to  each  other. 

Proof.  Since  AB  and  CD  are  in  the  same  plane,  if  they 
are  not  parallel  they  must  meet. 

If  they  do  meet  we  should  have  two  lines  drawn  through 
the  same  point  parallel  to  EF. 

This  contradicts  (?). 

Therefore  they  cannot  meet,  and,  by  definition  (§  107),  are 
parallel.  q.e.d. 

109.  Exercise.  If  a  line  be  drawn  on  this  page  parallel  to  the  upper 
edge,  show  that  it  is  also  parallel  to  the  lower  edge. 

110.  Exercise.  Give  an  example  of  two  lines  that  never  meet,  how 
far  soever  they  be  prolonged,  and  yet  are  not  parallel.  \_Note.  —  To  do 
this  the  student  must  leave  the  province  of  plane  geometry  and  think  of 
lines  in  different  planes.] 


BOOK  I  37 

Proposition  XVII.     Theorem 

111.  If  two  lines  are  perpendicular  to  the  same  line, 
they  are  parallel. 

E 


F 

Let  AB  and  CD  be  ±  to  ef. 

To  Prove  AB  and  CD  II  to  each  other. 

Proof.  If  AB  and  CD  are  not  parallel,  they  will  meet  at 
some  point.     (?) 

Then  we  should  have  two  perpendiculars  drawn  from  that 
point  to  EF. 

This  contradicts  (?). 

.-.  AB  and  CD  are  parallel.  q.e.d. 

112.    Problem.     Through   a   given  point   to   draw   a  line 

parallel  to  a  given  line.  r 

Let   P  be  the  given  point          \    ,/  ;               \    / 

and  AB  the  given  line.             D — }{^  •                  )(       E 

Required  to  draw  through  '     '^  I 

F  a  parallel  to  AB.  \  :c 

A ^-r^ ^^-r^ B 

Draw  PC  ±  to  AB.  ^~~r-"" 


Through  P  draw  D^  J_  to 
C. 
Prove  DE  and  AB  parallel,     q.e.f. 


PC.  '-Jy 


113.    Definitions.     A  straight  line  that  cuts  two  or  more 
lines  is  called  a  transversal. 


38  PLANE   GEOMETRY 

If  two  lines  are  cut  by  a  transversal,  eight  angles  are 
formed,  which  are  named  as  follows  : 

The  four  angles  [Zl,  Z2,  Z7,  and  Z8],  lying  without  the 
two  lines,  are  called  exterior  angles. 

The  four  angles  [Z3,  Z4,  Z.5, 
and  Z  6],  lying  within  the  two  lines, 
are  called  interior  angles. 

The  two  pairs  of  exterior  angles 
[Z  1  and  Z  7,  Z  2  and  Z  8],  lying  on 
the  same  side  of  the  transversal,  are 
called  exterior  angles  on  the  same  side. 

The  two  pairs  of  interior  angles 
[Z3  and  Z.6,  Z4  and  Z^,  lying 
on  the  same  side  of  the  transversal,  are  called  interior  angles 
on  the  same  side. 

The  four  pairs  of  angles  [Z 1  and  Z  5,  Z  2  and  Z  6,  Z  3  and 
Z  7,  Z4  and  Z8],  lying  on  the  same  side  of  the  transversal, 
one  an  exterior  and  the  other  an  interior  angle,  are  called 
corresponding  angles. 

The  two  pairs  of  exterior  angles  [Z  1  and  Z  8,  Z  2  and  Z  7], 
lying  on  opposite  sides  of  the  transversal,  are  called  alternate 
exterior  angles. 

The  two  pairs  of  interior  angles  [Z  3  and  Z  6,  Z  4  and  Z  5], 
lying  on  opposite  sides  of  the  transversal,  are  called  alternate 
interior  angles. 

The  four  pairs  of  angles  [Z  1  and  Z  6,  Z  2  and  Z  5,  Z  3  and 
Z  8,  Z  4  and  Z  7],  lying  on  opposite  sides  of  the  transversal, 
one  an  exterior  and  the  other  an  interior  angle,  are 
called  alternate  exterior  and  iyiterior 
angles. 

114.  Exercise.  Show  that  if  any  one 
of  the  following  sixteen  equations  is  true, 
the  other  fifteen  equations  are  also  true. 


7 


A 


BOOK  I  39 

1.  Z3  =  Z6.  9.  Z3  +  Z5  =  2R.A.'s. 

2.  Z4  =  Z5.  10.  Z4  +  Z6  =  2R.A.'s. 

3.  Z1=Z8.  11.  Zl +  Z7.=  2R.A.'s. 

4.  Z2=Z7.  12.  Z2  +  Z8  =  2  R.A.'s. 

5.  Z1=Z6.  13.  Zl +  Z6  =  2  R.A.'s. 

6.  Z2  =  Z6.  14.  Z2  +  Z5  =  2  R.A.'s. 

7.  Z3  =  Z7.  15  Z3  +  Z8  =  2  R.A.'s. 

8.  Z4  =  Z8.  16.  Z4  +  Z7=  2  R.A.'s. 

Proposition  XVIII.     Theorem 

115.   If  two  lines  are  cut  hy  a  transversal,  mahing  the 
alternate  interior  angles  equal,  the  lines  are  parallel. 


Let  AB  and  CD  be  cut  by  the  transversal  EF,  making 
Z1=Z2. 

To  Prove  AB  and  CD  parallel. 

Proof.  From  M,  the  middle  point  of  SO,  draw  MH  A-  to  CD, 
and  prolong  MH  until  it  meets  AB  in  some  point  G. 

Prove  the  A  GMO  and  MSH  equal  in  all  respects. 

Whence  Z.h  =  Ag. 

Z-H\s  by  construction  a  R.A. 

.-.  Z  (?  is  a  R.A. 
AB  and  CD  are  parallel.     (?)  q.e.d. 


40  PLANE   GEOMETRY 

116.  Corollary.  If  two  lines  are  cut  by  a  transversal^ 
making  any  one  of  the  following  six  cases  true,  the  lines  are 
parallel. 

1.  The  alternate  interior  angles  equal. 

2.  The  alternate  exterior  angles  equal. 

3.  The  corresponding  angles  equal. 

4.  The  sum  of  the  interior  angles  on  the  same  side  equal 
to  two  E.A.'s. 

5.  The  sum  of  the  exterior  angles  on  the  same  side  equal 
to  two  E-.A.'s. 

6.  The  sum  of  the  alternate  interior  and  exterior  angles 
equal  to  two  E-.A.'s. 

117.  Exercise.  FE  intersects  AB 
and  CD,  making  Z  m  =  f  R.A. 

What  value  must  /.n  have  in  order 
that  AB  and  CD  shall  be  i)arallel  ? 

118.  Exercise.  Through  a  given 
point  to  draw  a  parallel  to  a  given  line. 
(This  exercise  is  to  be  based  on  §  115. 
Another  solution  was  given  in  §  112.) 

[Through  the  given  point  P  draw  any 

line  PM  to  the  given  line  AB.    Through  in g 

P  draw    CD,  making   Z2  =  Z1.     Prove  ^ 
CD  parallel  to  AB. 

Work  this  exercise  by  making  the  alternate  exterior  angles  equal; 
also  by  making  the  corresponding  angles  equal.] 

119.  Exercise.  The  sum  of  two  angles  of  a  triangle  cannot  equal 
two  right  angles. 

120.  Exercise.  The  bisectors  of  the  equal  angles  1  and  2  in  the 
figure  of  §  118,  are  parallel. 


BOOK  1  41 

Proposition  XIX.     Theorem 

121.   If  two  parallels  are  cut  by  a  transversal,  the 
alternate  interior  angles  are  eqiial . 


Let  the  parallel  lines  AB  and  CB  be  cut  by  the  trans- 
versal EF. 

To  Prove  Aaos=Z.  OSD. 

Proof.     Suppose  Zaos  is  not  equal  to  Z  OSD. 

Draw  GH  through  0,  making  Z  GOS  =  Z  OSD. 

GH  and  CD  are  parallel.     (?) 

AB  and  CD  are  parallel.     (?) 

Through  0  there  are  two  parallels  to  CD,  which  contra- 
dicts (?). 

.*.  The  supposition  that  ZAOS  and  Z  OSD  are  unequal,  etc. 

Q.E.D. 

122.  Corollary  I.  If  two  parallels  are  cut  by  a  transversal, 
the  six  cases  of  §  116  are  true. 

123.  Corollary  II.  If  a  line  is  perpendicular  to  one  of 
two  parallels,  it  is  perpendicular  to  the  other  also. 

124.  Exercise,  The  bisectors  of  two  alternate  exterior  angles, 
formed  by  a  transversal  cutting  two  parallel  lines,  are  parallel. 

125.  Exercise.  If  a  line  joining  two  parallels  is  bisected,  any  other 
line  through  the  point  of  bisection,  and  joining  the  parallels,  is  also 
bisected. 

126.  Exercise.  If  AB  and  CD  are  parallel  (§  117),  and  Zw  =  If 
R.A.,  find  the  values  of  the  other  seven  angles. 


42  PLANE   GEOMETRY 

Proposition  XX.     Theorem 

127.  If  two  lines  are  cut  hy  a  transversal,  making  the 
sum  of  the  interior  angles  on  the  same  side  less  than  two 
right  angles,  the  lines  will  meet  if  sufficiently  produced. 


Let  AB  and  CD  be  cut  by  EF,  making  Z  1  4-  Z  2  <  2  R.  A.'s. 
To  Prove  that  AB  and  CD  will  meet. 

Proof.     If  AB  and  CD  do  not  meet,  they  are  parallel.     (?) 
If  they  are  parallel,  Z 1  +  Z  2  =  2  R.A.'s.     (?) 

This  contradicts  (?). 

.*.  they  cannot  be  parallel  and  must  meet.  Q.E.n 

128.  Corollary.  If  two  lines  are  cut  by  a  transversal,-  mak- 
ing any  one  of  the  six  cases  of  §  116  untrue,  the  lines  will  meet 
if  sufficiently  produced. 

129.  Exercise.  The  bisectors  of  any  two  exterior  angles  of  a  tri- 
angle will  meet. 

Prove  that  DA  and  FC  meet. 

Suggestion.     ZEAB<2  E.A.'s.    (?) 

ZKIR.A. 

Z3<1  R.A. 
Similarly,  Z4<1  R.A. 

Whence,      /3  +Z  4<2  R.A.'s. 


BOOK  I 


43 


130.  Definition.  Each  angle,  viewed  from 
its  vertex,  has  a  right  side  and  a  left  side. 

AB  is  the  right  side  of  /.ABC,  and  £C  is  its 
left  side. 


Proposition  XXI.     Theorem 

131.  If  two  angles  have  their  sides  parallel,  right  side 
to  right  side,  and  left  side  to  left  side,  the  angles  are 
equal. 

a 


Let  Z 1  and  Z.  2  have  their  sides  parallel,  right  side  to  right 
side,  and  left  side  to  left  side. 

To  Prove  Z1  =  Z2. 

Proof.     Prolong  AB  and  EF  until  they  intersect. 

Z 1  =  Z  3.     (?) 

Z3  =  Z2.     (?) 

Z  1  =  Z  2.       (?)  Q.E.D. 


44 


PLANE    GEOMETRY 


132.  Corollary.  If  two  angles  have  their  sides  parallel, 
right  side  to  left  side,  and  left  side  to  right  side,  the  angles  are 
supplemeyitary. 

Let  Zl  and  Z2  have  their 
sides  parallel,  right  side  to 
left  side  and  left  side  to  right 
side. 

To  Prove  Zl+Z  2=2  R.A.'s.  d— 

Proof.     Z  2  =  Z  3,  Z  1+Z  3=2  E.A.'s.  (?) 

.-.  ZH-Z2  =  2R.A.'s.  Q.E.D. 

133.  Exercise.     ZlandZ2  have  their  sides 
parallel,  right  side  to  right  side,  etc. 

Z  2  and  Z.  3  have  their  sides  parallel,  right  side 
to  right,  etc.     Prove  that  Z 1  =  Z  3. 

Proposition  XXII.     Theorem 

134.  If  the  sides  of  one  angle  are  perpendicular  to  those 
of  another,  right  side  to  right  side  and  left  side  to  left 
side,  the  angles  are  equal. 

J  E 

H 


Let  Z  1  and  Z  2  have  DE  ±  to  BC  and  FE  A.  to  AB. 
To  Prove  Z 1  =  Z  2. 

Proof.     Draw  BH  II  to  ED  and  BJ  II  to  FE.     Z  3  =  Z  2.  (?) 

BE  \^  1.  to  BC  (?)  and  JB  is  .L  to  AB.  (?) 

Z3  +  Z4  =  1R.A.  and  Zl+Z4  =  l  R.A. 

Z  3  =  Z  1.     (?)  .-.  Z  2  =  Z 1.     (?)        Q.E.D. 


BOOK  I 


45 


135.    Corollary.     If  the 


of  one  angle  are  perpendicular 


to  those  of  another,  right  side  to  left 
side  and  left  side  to  right  side,  the 
angles  are  supplementary. 

To  Prove  Z  1  +  Z  2  =  2  E.A.'s. 
Proof.     Prolong  AB  to  G. 
Show  that   Z3  =  Z2. 

Zl  +  Z3==2E.A.'s. 
.-.  Z  l-f  Z  2  =  2  E.A.'s. 

136.  Exercise.  In  A^JSC,  AD  is 
1.  to  BC  and  CE  ±  to  AB.  Compare 
Zl  and  Z2. 

137.  Definition.    Two  triangles 
are  mutually  equiangular  when  ths  angles  of  one  are  equal  re- 
spectively to  the  angles  of  the  other. 


Proposition  XXIII.     Theorem 

138.  The  Slim  of  the  interior  angles  of  a  triangle  is  two 
right  angles. 


Let  ABC  be  any  A. 

To  Prove  Zl+Z2-hZ3  =  2  R. A.'s. 

Proof o     Draw  DE  through  the  vertex  B,  parallel  to  AC. 
Z  4  =  Z  2  and  Z  5  =  Z  3.     (?) 

Z44-Z1  +  Z5  =  2  R.A.'s.     (?) 

Z2  +  Zl+Z3  =  2R.A.'s.     (?)  Q.E.D. 


46  PLANE   GEOMETRY 

139.  Corollary  I.  If  two  aiigles  of  a  triangle  are  knowrij 
the  third  can  be  found  by  subtracting  their  sum  from  two  right 
angles. 

140.  Corollary  II.  If  two  angles  of  one  triangle  are  equal 
respectively  to  two  angles  of  another,  the  third  angles  are  equal, 
and  the  triangles  are  mutually  equiangular. 

141.  Corollary  III.  A  triangle  can  contain  only  one  right 
angle;  and  it  can  contain  only  one  obtuse  angle. 

142.  Corollary  IV.  In  a  right-angled  triangle,  the  sum  of 
the  acute  angles  is  one  right  angle. 

143.  Corollary  V.  Since  an  equilateral  triangle  is  also 
equiangular,  each  angle  is  two  thirds  of  a  right  angle. 

144.  Corollary  VI.  An  exterior  angle  of  a  triangle  (formed 
by  prolongiyig  a  side)  is  equal  to  the  sum  of  the  two  opposite  interior 
angles  of  the  triangle. 

145.  Exercise.  One  of  the  acute  angles  of  a  R.A.  A  is  f  R.A. 
What  is  the  other  ? 

146.  Exercise.  Find  the  angles  of  a  A,  if  the  second  is  twice  the 
first,  and  the  third  is  three  times  the  second. 

147.  Exercise.  Find  the  angles  of  an  isosceles  A,  if  a  base  angle  is 
one  half  the  vertical  angle. 

148.  Exercise.     Given  two  angles  of  a  triangle,  construct  the  third. 

149.  Exercise.  Prove  that  the  bisectors  of  the  acute  angles  of  an 
isosceles  right-angled  triangle  make  with  each  other  an  angle  equal  to 
H  R.A.'s. 

150.  Exercise.  Prove  that  the  bisector  of  an  exterior  vertical  angle 
of  an  isosceles  triangle  is  parallel  to  the  base. 


BOOK  I  47 

151.   Exercise.      Prove  §  138,  using  these  figures. 


152.  Definitions.    A  portion  of  a  plane  bounded  by  straight 
lines  is  called  a  polygon. 

The  bounding  line  of  a  polygon  is  its  perimeter. 

A  diagonal  of  a  polygon  is  a  straight 
line  joining  any  two  of  its  vertices  that 
are  not  consecutive. 

A  three-sided  polygon  is  a  triangle;  a 
four-sided  polygon  is  a  quadrilateral;  a 
five-sided  polygon  is  a  pentagon;  a  six- 
sided  polygon  is  a  hexagon;  an  eight-sided  polygon  is  an 
octago7i;  a  ten-sided  polygon  is  a  decagon;  and  a  fifteen-sided 
polygon  is  a.  pentedecagon. 

A  polygon  whose  angles  are  equal  is  an  equiangular  polygon. 

A  polygon  whose  sides  are  equal  is  an  equilateral  polygon. 

A  polygon  that  is  both  equilateral  and  equiangular  is  a 
regular  polygon. 

153.  Exercise.     Show  that  an  equilateral  triangle  is  regular. 

154.  Exercise.     Show,  by  drawings,  that  an  equilateral  quadrilateral 
is  not  necessarily  regular. 

155.  Exercise.     How  many  diagonals  can  be  drawn  in  a  triangle  ? 
In  a  quadrilateral  ?     In  a  hexagon  ? 

156.  Exercise.     How  many  diagonals  can  be  drawn  from  one  vertex 
in  a  polygon  of  n  sides  ?    How  many  from  all  the  vertices  ? 


48  PLANE   GEOMETRY 

Proposition  XXIV.      Theorem 

157.  Tixe  sum  of  the  interior  angles  of  a  polygon  is 
twice  as  many  right  angles  as  tJie  polygon  has  sides,  less 
four  right  angles 


Let  ABC  ...  J^'  be  a  polygon  of  n  sides. 

To  Prove  that  the  sum  of  its  interior  angles  is  (2  n— 4)  E.A.'s. 

Proof.    From  any  point  within  the  polygon,  as  0,  draw  lines 

to  all  the  vertices. 

The  polygon  is  now  divided  into  n  A.     (?) 

The  sum  of  the  angles  of  each  A  is  2  E.A.'s.     (?) 

The  sum  of  the  angles  of  the  w  A  is  2  w  R.  A.'s.     (?) 

The  sum  of  the  angles  of  the  polygon  is  equal  to  the  sum  of 

the  angles  of  the  A,  diminished  by  the  sum  of  the  angles 

about  O ;  that  is,  by  4  R.A.'s. 

.-.  the  sum  of  the  angles  of  the  polygon  is  (2  n  —  4)  E.A.'s. 

Q.E.D. 

158.  Corollary.  The  value  of  each  angle  of  an  equiangular 
polygon  of  n  sides  is  M.AJs. 

159.  Exercise.  What  is  the  sum  of  the  interior  angles  of  a  quadri- 
lateral ?    Of  a  pentagon  ?    Of  a  hexagon  ?    Of  a  polygon  of  100  sides  ? 

160.  Exercise.  How  many  sides  has  the  polygon  in  which  the  sum 
of  the  interior  angles  is  20  R.A.'s  ?  26  R.A.'s  ?  98  R.A.'s  ?  (2  s  -  4) 
R.A.'s? 


BOOK   I 


49 


161.  Exercise.     How  many  sides  has  the  equiangular  polygon  in 
which  one  angle  is  f  R.A.  ?     1  R.A.  ?     If  R.A.  ?     114  R.A.  ? 

162.  Exercise.     How  many  sides  has  the  equiangular  polygon  in 

which  the  sum  of  four  angles  is  6  R.A.'s  ? 

C  D 

163.  Exercise.  Prove  §  157,  using 
this  figure.  Show  that  the  polygon  is 
divided  into  w  —  2  triangles,  the  sum  of 
the  angles  of  which  is  equal  to  the  sum  of 
the  angles  of  the  polygon. 


Proposition  XXV.     Theorem 

164.  The  sum  of  the  exterior  angles  of  a  polygon,  formed 
hy  prolonging  one  side  at  eaeh  vertex,  is  four  R.A.'s. 


Let  ^5  ...  ^  be  a  polygon  of  n  sides. 

To  Prove  that  the  sum  of  its  exterior  angles  1,  2,  3,  etc.,  is  4 

E.A.'s. 

Proof.     The  sum  of  each  exterior  angle  and  its  adjacent 

interior  angle  is  2  R.A.'s.     (?) 

2n  R.A.'s  is  the  sum  of  all  exterior  and  interior  angles.     (?) 

(2  n  —  4)  R.A.'s  is  the  sum  of  the  interior  angles.     (?) 

4  R.A.'s  is  the  sum  of  the  exterior  angles.     (?)  q.e.d. 

SANDERS'    GEOM.  —  4 


60 


PLANE   GEOMETRY 


165.  Scholium.  It  is  indifferent  which  side  is  prolonged 
at  any  vertex,  as  the  exterior  angles  formed  at  any  vertex  by 
prolonging  both  sides  are  equal. 

166.  Exercise.  How  many  sides  has  the  polygon  in  which  the  sum 
of  the  interior  angles  is  five  times  the  sum  of  the  exterior  angles  ? 

167.  Exercise.  Complete  the  following  table.  The  polygons  are 
equiangular. 


No.  of  Sides. 

Value  of  each  Interior 
Angle. 

Value  of  each  Exterior 
Angle. 

3 
4 
6 

12 

f  R.A. 
IR.A. 

If  R.A. 

f  R.A. 
IR.A. 

|R.A. 

Proposition  XXVI.     Theorem 
168.   The  sum  of  two  sides  of  a  triangle  is  greater  than 
the  third  side,  hut  the  difference  of  two  sides  of  a  triangle 
is  less  than  the  third  side. 


Let  ABC  be  any  A. 

To  Prove  AB  +  bOAC, 

Proof.     Apply  axiom  14. 
Let  BEF  be  any  A. 

To  Prove  BE  —  EF<  BF. 

Proof.  BE<BF-{-EF.       (?) 

Subtract  EF  from  both  members.      (?)  . 
BE—EF<BF, 


Q.E.D. 


Q.E.D 


BOOK  I  51 

169.  Exercise.     Can  a  triangle  have  for  its  sides  6  in.,  7  in.,  and 
15  in.  ? 

170.  Exercise.     Two  sides  of  a  triangle  are  5  ft.  and  7  ft.    Between 

what  limits  must  the  third  side  lie  ? 

171.  Exercise.     Each  side  of    a  triangle  is  less  than    the    semi- 
perimeter. 

B 

172.  Exercise.    The  sum  of  the  lines  drawn 

from  a  point  within  a  triangle  to  the  three  ver- 
tices is  greater  than  the  semi-perimeter. 

'FvoyqOA-\-OB-^OC>UAB-\-BC-\-CA\      X-^"^^ 

k^^ ^C 

173.  Definition.     A  medial  line  of  a 

triangle  (or  simply  a  median)  is  a  line  drawn  from  any  vertex 
of  the  triangle  to  the  middle  point  of  the  opposite  side. 

174.  Exercise.    A  median  to  one  side  of  a  triangle  is  less  than  one 
half  the  sum  of  the  other  two  sides. 

To  prove  BD  <i  {AB  -\-BC).  ^ 

Prolong  BD  until  DE  =  BD. 

Draw  CE. 

Prove   Aabd    and  DCE    equal,  ^ 
whence  EC  =  AB. 

BC  -\-  CE>BE.      (?) 

Divide  both  members  by  2,  recol-  E 

lecting  that  BD  =  DE  and  EG  =  AB. 

175.  Exercise.     The  sum  of  the  three  medians  of  a  triangle  is  less 
than  its  perimeter. 

Suggestion.     Use  the  preceding  exercise. 

A^: i:^^B 

176.  Exercise.     The    lines   AB  and 

CD  have  their  extremities  joined  by  CB 
and  AD.  C 

Prove  CB  +  AD>AB-\-CD. 


62 


PLANE   GEOMETRY 


Proposition  XXVII.     Theorem 

177.  //  froiiv  a  point  within  a  triangle  two  lines  are 
drawn  to  the  ejctremities  of  a  side,  their  sum  is  less  than 
that  of  the  two  remaining  sides  of  the  triangle. 


Let    ABC  be  any  A,   O  any  point  within,  and  OA  and  OC 
lines  drawn  to  the  extremities  of  A  C. 

To  Prove  OA  +  OC<AB  +  BC, 

Proof.     Prolong  AO  to  D. 

AB-\-BD>AO-\-OD.      (?) 

OD-{-DC>OC.      (?) 

Add  these  inequalities  and  show  that  AB  -\-BC>AO  -\-  OC. 

Q.E.D. 

178.  Exercise.     FroYe  ZA0C>  Z  ABC. 

Suggestion.     Show  that  ZAOC>ZODC  and  Z  ODC>  ZABC.     Give 
another  proof  for  this  exercise  without  prolonging  AO. 

179.  Exercise.     The  sum  of  the  Hues  drawn  from  a  point  within  a 
triangle  to  the  three  vertices  is  less  than  the  perimeter  of  the  triangle. 


180.   Exercise.     Prove  that  the  perimeter  of  the 
star  is  greater  than  that  of  the  polygon  ABCDEF. 


BOOK  1 


53 


Proposition   XXVIII.     Theorem 

181.  If  two  triangles  have  two  sides  of  the  one  equal 
respectively  to  two  sides  of  the  other,  and  the  included 
angles  unequal,  the  third  sides  are  unequal,  and  the 
greater  third  side  belongs  to  the  triangle  having  the 
greater  included  angle. 


Let 

and 
To  Prove 


C   D 


the  A  ABC  and  DBF  have 
ab  =  de,  bc  =  ef 
Zb>Z.e. 

AC>DF. 


Proof.     Of  the  two  sides,  AB  and  BC,  let  AB  be  the  one 
which  is  not  the  larger. 

Draw  B  G,  making  Zabg=ZF;  prolong  B  G,  making  BG= EF. 

Draw  AG. 

Prove  Aabg  =  A DEF,  whence  AG  =  DF. 

Draw  BH  bisecting  Z  GBC. 

Draw  GH. 

Prove  Agbh  =  Ahbc.     Whence  HG=HC. 

AH+HG>AG.      (?) 

AOAG.  (?) 

AC>J)F.  (?)  Q.E.D 


54 


PLANE   GEOMETRY 


182.  Converse.  If  two  triangles  have  two  sides  of  tlie 
one  equal  respectively  to  two  sides  of  the  other,  and  the  third 
sides  unequal,  the  included  angles  are  unequal,  and  the  greater 
included  angle  belongs  to  the  triangle  having  the  greater 
third  side. 


Let  A  ABC  and  DEF  have 

AB  =  DE,  BG  =  EF, 

and  AC>  DF. 

To  Prove  Zb>E. 

Proof.         Zb=  Ze,  ov  Zb<Ze,  or  Zb>Z.E. 

Show  that  Z  B  cannot  equal  Z  E. 
Show  that  Z  B  cannot  be  less  than  Z  E. 

.-.  Zb>Ze. 


Q.E.D. 


183.  Exercise.  B  is  fifty  miles  west  of  A.  C  is  forty  miles  north 
of  B,  and  D  is  forty  miles  southeast  of  B.  Show  that  O  is  a  greater 
distance  from  A  than  D  is. 


184.  Exercise.  In  the  isosceles  triangle  ABC, 
BD  is  drawn  to  a  point  D  on  the  base  AG  s,o  that 
AD>DC. 

Prove  Z  ADB  >ZBDC. 

Suggestion.  Compare  A  ABD  and  DBC,  using 
§  182.  Then  compare  A  ADB  and  DBC,  using 
§144. 


BOOK  I 


55 


Proposition  XXIX.     Theorem 

185.   //  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  them  are  equal. 


Let  ^5C  be  a  A  having  Z .4  =  Z  C. 

To  Prove  AB=BC. 

Proof.     Draw  BD  bisecting  Zb. 

Prove  ^ABD  and  BDG  mutually  equiangular. 

Prove  Aabd  and  BDC  equal  in  all  respects. 

Whence  AB=BC.  q.e.d. 

186.  Corollary.  An  equiangular  tri- 
angle is  equilateral. 

187.  Exercise.  ABC  is  an  isosceles  triangle 
hSiYmgAB  =  BC. 

AD  and  DC  bisect  Z  A  and  Z  C  respectively. 

Prove  AD  =  DC. 

188.  Exercise.  If  the  bisector  of  an  angle  of 
a  triangle  bisects  the  opposite  side,  it  is  also  per- 
pendicular to  that  side,  and  the  triangle  is  isosceles. 

Let  BB  bisect  Z  B  and  also  bisect  AC. 
To  Prove  BD  ±  to  AC,  and  A  ^5(7  isosceles. 
Suggestion.     Prolong  BD  until  DE  =  BD. 
Prove  A  ABD  =  ADEC. 

Whence  Zl  =  Z6. 
Prove  IlBCE  isosceles. 


56 


PLANE   GEOMETRY 


Proposition  XXX.     Theorem 

189.  If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  to  them  are  unequal,  the  greater  angle  being 
opposite  the  greater  side;  and  conversely,  if  two  angles 
of  a  triangle  are  unequal,  the  sides  opposite  them  are 
unequal,  the  greater  side  lying  opposite  the  greater  angle. 


Let  ^^C  be  a  A  having  AB>BG. 

To  Prove  ZOZ.A. 

Proof.  On  AB  lay  off  BD=BG  and  draw  DC. 
Z1  =  Z2.  (?) 
Z1>Z^.  (?) 
Z2>Z^.  (?) 
Ac>Aa.  (?) 
Let  je:2?'(?  be  a  A  having  Zg'>Zjs;. 

To  Prove  EF::^FG. 

Proof.     Draw  CD  making  Z 1  =  Z  J?. 

HG+HF>FG.      (?) 
HG  =  EH.  (?) 

EF  >  FG.  (?) 


Q.E.D. 


Q.E.D. 


190.  Exercise.     Prove  the  converse  to  this  proposition  indirectly. 
Show  that  EF  can  neither  be  equal  to  FG  nor  less  than  FG,  and  must 

consequently  be  greater  than  FG. 

191.  Exercise.     ABC  is  a  triangle 
having  ^O^C 

AD  bisects  A  A  and  BD  bisects  Z.  B. 

Prove  AD>BD. 


BOOK  I 


57 


Proposition  XXXI.     Theorem 

192.  If  two  right-angled  triangles  have  the  hypote- 
nuse and  a  side  of  one  equal  respectively  to  the  hypote- 
nuse and  a  side  of  the  other,  the  triangles  are  equal  in 

all  respects. 

S  E 


Let  ABC  and  DEF  be  two  R.A.A  having  hypotenuse  ab=z 
hypotenuse  DE,  and  AC  =  DF. 

To  Prove  the  A  ABC  and  DEF  equal  in  all  respects. 

Proof.  Place  A  ABC  so  that  AC  coincides  with  its  equal  DF, 
A  falling  at  D,  and  C  at  F,  and  the  vertex  B  falling  at  some 
point  G  on  the  opposite  side  of  the  base  DF  from  E. 

Show  that  EF  and  FG  form  a  straight  line. 

Show  (in  the  Agde)  that   Z.G  =/.E. 
Z3  =Z4.     (?) 

Adfg  and  DFE  are  equal  in  all  respects.     (?) 

A  ABC  and  DFE  are  equal  in  all  respects.  q.e.d. 

193.  Exercise.  If  a  line  is  drawn  from  the  vertex  of  an  isosceles 
triangle  ±  to  the  base,  it  bisects  the  base  and  the  vertical  angle. 

194.  Definitions.  A  quadrilateral  having  its  opposite  sides 
parallel  is  called  a  parallelogram. 

A  quadrilateral  with  one  pair  of  parallel  sides  is  a  trapezoid. 


58  PLANE   GEOMETRY 

A  quadrilateral  with  no  two  of  its  sides  parallel  is  a 
trapezium. 

A  parallelogram  whose  angles  are  right  angles  is  a  rectangle. 

A  parallelogram  whose  angles  are  oblique  angles  is  a  rhomboid. 

A  square  is  an  equilateral  rectangle;  and  a  rhombus  is  an 
equilateral  rhomboid. 

Proposition  XXXII.     Theorem 

195.  The  opposite  sides  of  a  parallelogram  are  equal; 
and  conversely,  if  the  opposite  sides  of  a  quadrilateral 
are  equal,  the  figure  is  a  parallelogram. 

Fi 


Let  ABGB  be  a  parallelogram. 

To  prove  ab  =  CB  and  BC  =  AB, 

Proof.     Draw  the  diagonal  BB. 

Z  1  =  Z  2.     (?) 

Z3  =  Z4.     (?) 
Show  that  A  abb  =Abcb. 

Whence  ab  =  CB  and  BC  =  AB.  q.e.d. 

Let  EFGH  be  a  quadrilateral  having  EF^  GH  and  FG  =  EH. 
To  prove  EFGH  a  parallelogram. 
Proof.     Draw  the  diagonal  FH. 
Prove  A  EFH  =  A  FGH. 

Whence  Z 1  =  Z2  and  Z 3  =  Z4. 

Since  Zl  =  Z2,  FG  and  EH  are  parallel.     (?) 

Similarly  EF  is  parallel  to  GH. 

EFGH  is  a  parallelogram.  q.e.d. 


BOOK  I  59 

196.  Corollary  I.  xi  diagonal  of  a  parallelogram  divides  it 
into  two  triangles  equal  in  all  respects. 

197.  Corollary  II.  Two  parallelograms  are  equal  if  they 
have  two  adjacent  sides  and  the  included  angle  of  one  equal  respec- 
tively to  two  adjacent  sides  and  the  included  angle  of  the  other. 

198.  Corollary  III.  Parallels  included  between  two  paral- 
lels and  limited  by  them,  are  equal. 

Proposition  XXXIII.     Theorem 

199.  The  opposite  angles  of  a  parallelogram  are  equal ; 
and  conversely,  if  the  opposite  angles  of  a  quadrilateral 
are  equal,  the  figure  is  a  parallelogram. 

B< rC 


Let  ABCD  be  a  parallelogram. 

To  Prove  Z  ^  =  Z  C  and  /.B  =  Z.D. 

Proof.     Show  by  §  131  that  Aa=/.C  and  Zb=Z.D. 

Q.E.D. 

Conversely.     In   the    quadrilateral   ABCD    let   /.  A  = /.  C 
and  Z.B  =  Z.D. 

To  Prove  ABCD  a  parallelogram. 

Proof.       Z^-hZj5+Zc  +  Zi)  =  4  K.  A.'s.     (?) 

Z  JL  =  Z  c  and  Z  £  =  Z  D. 
2Z^  +  2Z5=4R.A.'s.     (?) 

Z  ^  +  Z  i5  =  2  R.  A.'s.     (?) 
BG  and  AD  are  parallel.     (?) 
Similarly  prove  AB  and  CD  parallel. 
ABCD  is  a  parallelogram.     (?)  q.e.d. 


60  PLANE   GEOMETRY 

200.  Corollary.  Tlie  adjacent  angles  of  a  parallelogram  are 
supplementai'y  ;  and  conversely,  if  the  adjacent  angles  of  a  quadn- 
lateral  are  supplementary,  the  figure  is  a  parallelogram. 

201.  Exercise.  If  one  of  the  angles  of  a  parallelogram  is  a  right  angle, 
the  other  three  are  also  right  angles. 

202.  Exercise.  If  one  angle  of  a  parallelogram  is  f  R.A.,  how  large 
are  the  others  ? 

203.  Exercise.  If  two  sides  of  a  quadrilateral  are  parallel,  and  a 
pair  of  opposite  angles  are  equal,  the  figure  is  a  parallelogram. 

204.  Exercise.  If  an  angle  in  one  parallelogram  is  equal  to  an  angle 
in  another,  the  remaining  angles  are  equal  each  to  each. 

Proposition  XXXIV.     Theorem 

205.  If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  the  figure  is  a  parallelogram. 


\ 


Let  ABCD  be  a  quadrilateral  having  BC  and  AD  equal  and 
parallel. 

To  Prove  ABCD  o,  parallelogram. 
Proof.     Draw  the  diagonal  BD. 

AABD  =ABCD.      (?) 
Whence  AB  =  CD. 

Prove  ABCD  Si  parallelogram.     [§  195.     Converse.]       q.e.d. 

206.  Exercise.  The  line  joining  the  middle  points  of  two  opposite 
sides  of  a  parallelogram  is  parallel  to  each  of  the  other  two  sides  and 
equal  to  either  of  them. 


BOOK  I 


61 


Proposition  XXXV.     Theorem 

207.  The  diagonals  of  a  parallelogram  bisect  each 
other;  and  conversely ^  if  the  diagonals  of  a  quadri- 
lateral bisect  each  other,  the  figure  is  a  parallelogram. 


Let  ABCB  be  a  parallelogram,  DB  and  AC  its  diagonals. 

To  Prove  BO  =  OD  and  AO  =  OC. 

Proof.    I'rove  Aboc=AAOD,  whence  BO  =  OD  and  A 0 

Conversely.     In  the  quadrilateral  ABCD, 
Let  AO  =  OC  and  BO  =  OB. 

To  Prove  ABCD  a  parallelogram. 
Proof.     Prove  Aboc  =AA  OD, 

whence  Z  1  =  Z 2  and  BC  =  AD, 

Prove  ABCD  a  parallelogram.     (§  205.) 

208.  Corollary  I.     The  diagonals  of  a  square   b 

1.  Are  equal. 

2.  Bisect  each  other. 

3.  Are  perpendicular  to  each  other. 

4.  Bisect  the  angles  of  the  square. 

209.  Corollary  II.     The  diagonals  of  a  rhombus 

1.  Are  unequal.  B 

2.  Bisect  each  other. 

3.  Are  perpendicular  to  each  other. 

4.  Bisect  the  angles  of  the  rhombus. 
To  prove  the  diagonals  unequal. 


=  0C. 

Q.E.D. 


62  PLANE   GEOMETRY 

first  show  that  Z  A   and  Z  D   of   the  rhombus  are  unequaL 
(They  are  supplementary  and  oblique.) 
Then  apply  §  181  to  A  ABD  and  ^CD. 

210.  Corollary  III.  The  diagonals  of  a  rectangle  that  is 
not  a  square 

1.  Are  equal.  ^ 

2.  Bisect  each  other. 

3.  Are  not  perpendicular  to  each  other. 

4.  Do  not  bisect  the  angles  of  the  rec- 
tangle. 

To  prove  that  the  diagonals  are  not  perpendicular  to  each 
other,  apply  §  182  to  Aboc  and  COD.  (BC  and  CB  are 
unequal  because  the  rectangle  is  not  a  square.) 

To  prove  that  the  diagonals  do  not  bisect  the  angles  of  the 
rectangle,  show  that  A  4  and  5  of  Aacb  are  unequal,  but 
Z  3  =  Z4.  (?)     .'.  Z  3  and  Z 5  are  unequal. 

211.  Corollary  IY.  The  diagonals  of  a  rhomboid  that  is 
not  a  rhombus 

1.  Are  unequal. 

2.  Bisect  each  other. 

3.  Are   not  perpendicular   to   each 
other. 

4.  Do  not  bisect  the  angles  of  the    ^i 
rhomboid. 

212.  Exercise.  Any  line  drawn  through  the  point  of  intersection  of 
the  diagonals  of  a  parallelogram  and  limited  by  the  sides  is  bisected  at  the 
point. 

213.  Exercise.  If  the  diagonals  of  a  parallelogram  are  equal,  the 
figure  is  a  rectangle. 

214.  Exercise.     Given  a  diagonal,  construct  a  square. 

215.  Exercise.  Given  the  diagonals  of  a  rhombus,  construct  the 
rhombus. 


BOOK  I 


63 


Proposition  XXXYI.     Theorem 

216.  If  from  a  point  without  a  line  a  perpendicular  is 
drawn  to  the  line,  and  oblique  lines  are  drawn  to  differ- 
ent points  of  it, 

I.  The  perpendicular  is  shorter  than  any  oblique  line. 

II.  Two  oblique  lines  that  ineet  the  given  line  at  points 
equally  distant  from  the  foot  of  the  perpendicular  are 
equal. 

III.  Of  two  oblique  lines  that  meet  the  given  line  at 
points  unequally  distant  from  the  foot  of  the  perpen- 
dicular j  the  one  at  the  greater  distance  is  the  longer. 


D  C  E  F 

I.  Let  ABhQ  the  given  line  and  P  the  point  without,  PC  the 
±,  and  PD  any  oblique  line. 

To  Prove  PC<PD. 

Suggestion.     Apply  §  189,  converse,  to  A  PCD. 

II.  Let  PD  and  PE  be  oblique  lines  meeting  AB  at  points 
equally  distant  from  C. 

To  Prove  PD  =  PE. 

III.  Let  PF  and  PD  be  oblique  lines,  F  being  at  a  greater 
distance  from  C  than  is  the  point  D. 

To  Prove  pf>pd. 

Suggestion.     Show  that  Z 1  is  obtuse.     Then  apply  §  189,  converse,  to 
APEF,  recollecting  that  PE  -  PD. 


64  PLANE   GEOMETRY 

217.  Corollary  I.  The  perpendicular  is  the  shortest  distance 
from  a  point  to  a  line,  and  conversely. 

218.  Corollary  II.  From  a  point  without  a  line  only  two 
equal  lines  can  he  drawn  to  the  line. 

Note.  The  number  of  pairs  of  equal  lines  that  can  be  drawn  from  a 
point  to  a  line  is  of  course  infinite. 

219.  Corollary  III.  If  from  a  point  without  a  line  a  per- 
pendicular and  two  equal  oblique  lines  he  drawn,  the  ohlique  lines 
meet  the  given  line  at  points  equally  distant  from  the  foot  of  the 
perpendicular. 

Suggestion.     Use  §  192. 

220.  Definition.  An  altitude  of  a  triangle  is  a  perpendic- 
ular drawn  from  the  vertex  of  any  angle  to  the  opposite  side. 

221.  Exercise.  The  sum  of  the  altitudes  of  a  triangle  is  less  than 
the  perimeter. 

Proposition  XXXYII.     Theorem 

222.  Two  parallels  are  everywhere  equally  distant. 


A- 


E F 


\ B 


"^  G  H 

Let  AB  and  CD  be  two  ll's. 

To  Prove  that  they  are  everywhere  equally  distant. 

Proof.  From  any  two  points  on  AB,  as  E  and  F,  draw  EG 
and  FH  ±  to  CD. 

They  are  also  ±  to  ^5  (?),  and  they  measure  the  distance 
between  the  parallels  at  E  and  F. 

EG  and  FH  are  parallel.     (?) 

EG  and  FH  are  equal.     (?) 

Therefore  the  parallels  are  equally  distant  at  E  and  F. 

Since  E  and  F  are  any  points  on  AB,  the  parallels  are  every- 
where equally  distant.  q.e.d. 


BOOK  I  65 

223.  Scholium.  The  term  distance  in  geometry  means 
shortest  distance. 

The  distance  from  one  point  to  another  is  measured  on  the 
straight  line  joining  them.     (Axiom  14.) 

The  distance  from  a  point  to  a  line  is  the  perpendicular 
drawn  from  that  point  to  the  line.     (§  216.) 

The  distance  between  two  parallels  is  measured  on  a  line 
perpendicular  to  both.     (§  222.) 

The  distance  between  two  lines  in  the  same  plane  that  are 
not  parallel  is  zero ;  for  distaiice  means  shortest  distance,  and  the 
lines  will  meet  if  sufficiently  produced. 

224.  Corollary.  If  two  points  are  on  the  same  side  of  a 
given  line  and  equally  distant  from  it,  the  line  joining  the  points  is 
parallel  to  the  given  line. 

225.  Exercise.  If  the  two  angles  at  the  B  C 
extremities  of  one  base  of  a  trapezoid  are  /  \ 
equal,  the  two  non-parallel  sides  are  equal.               /  |  •  \ 

Suggestion.     Draw  BE  and  CF  ±  to  AD.       j     \  I    \ 

BE=CF  (?).     Prove  k.  ABE  and  CDF     L — i i — 1 

equal.     Whence  AB  =  CD.  ^      ^  ^ 

226.  Exercise.  11  the  two  non-parallel  sides  of  a  trapezoid  are 
equal,  the  angles  at  the  extremities  of  either  base  are  equal. 

Suggestion.  In  the  figure  of  the  preceding  exercise,  prove  A  ABE 
and  CFD  equal.     Whence  ZA  =  ZD. 

227.  Exercise.  If  a  quadrilateral  has  one  pair  of  opposite  sides 
equal  and  not  parallel,  and  the  angles  made  by  these  sides  with  the  base 
equal,  the  quadrilateral  is  a  trapezoid. 

Suggestion.     In  the  figure  of  §  225,  let  AB  =  CD  and  ZA=  ZD. 
Prove  /S^  ABE  and  CFD  equal,  and  then  use  §  224. 

228.  Exercise.  If  two  points  are  on  opposite  sides  of  a  line,  and 
are  equally  distant  from  the  line,  the  line  joining  them  is  bisected  by  the 
given  line. 

229.  Exercise.  If  a  rectangle  and  a  rhomboid  have  equal  bases  and 
equal  altitudes,  the  perimeter  of  the  rectangle  is  less  than  that  of  the 
rhomboid. 

SANDERS'    GEOM. 6 


66 


PLANE    GEOMETRY 


Proposition  XXXVIII.     Theorem 

230.  Anj/  point  on  the  bisector  of  an  angle  is  equally 
distant  from  the  sides  of  the  angle;  and  any  point  not 
on  the  bisector  is  unequally  distant  from  the  sides. 


Let  ABC  be  any  angle,  BD  its  bisector,  and  P  any  point  on  BD. 

To  Prove  P  equally  distant  from  AB  and  BC. 

Proof.  Draw  PE  and  PF  perpendicular  to  AB  and  BC  respec- 
tively. 

Prove  A  EPB  =  A  PBF. 

Whence  PE  =  PF.  q.e.d. 

Let  ABC  be  any  angle,  BD  its  bisector,  and  P  any  point  with- 
out BD. 

To  Prove  P  unequally  distant  from  AB  and  BC. 

Proof.     Draw  PE  and  P//  ±  to  AB  and  BC  respectively. 

From  F  (where  PE  intersects  BD)  draw  FG  ±  to  BC. 

Draw./"i. 

FP  -\-  FG>  PG.      (?) 

PG  >  PH.  (?) 

FP  +  ^'G  >  PH.  (?) 

FE  =  FG.  (?) 

FP  -\-  FE  >  PH.  (?) 

PE  >  PH.  (?)  Q.E.J 


BOOK  I 


67 


231.  Corollary.     Any  point  that  is  equally  distant  from  the 
sides  of  an  angle  is  on  the  bisector. 

232.  Exercise.     Prove  the  second  part  of  §  230  indirectly. 
Suppose  PE=  PG.     Draw  PB.  q 

Prove  A  PEB  =  A  PBG. 

Whence  Z  PBE  =  Z  PBG. 

.:  PB  must  bisect  ZABC. 


Definition^.  The  locus  of  a  point 
satisfying  a  certain  condition  is  the  line, 
lines,  or  part  of  a  line  to  which  it  is  thereby 
restricted ;  provided,  however,  that  the  con- 
dition is  satisfied  by  every  point  of  such  line  or  lines,  and  by 
no  other  point. 

The  bisector  of  an  angle  is  the  locus  of  points  that  are  equally 
distant  from  its  sides;  for  by  §  230,  all  the  points  on  the 
bisector  are  equally  distant  from  the  sides,  and  all  points  with- 
out the  bisector  are  unequally  distant  from  the  sides. 

234.  Exercise.  What  is  the  locus  of  points  that  are  equally  distant 
from  a  given  point  ?     From  two  given  points  ? 

235.  Exercise.  What  is  the  locus  of  points  that  are  equally  distant 
from  a  given  line  ? 

236.  Exercise.  What  is  the  locus  of  points  that  are  equally  distant 
from  a  given  circumference  ? 

237.  Exercise.  The  bisectors  of  the  interior  angles  of  a  triangle  meet 
in  a  common  point. 

To  Prove  that  the  bisectors  AD,  BF,  and 
EC  meet  <n  a  common  point. 

Prove  tiiat  AD  and  EC  meet.    (§  127. ) 

Call  their  point  of  meeting  0. 

0  is  equally  distant  from  AB  and  AC.   (?) 

0  is  equally  distant  f rom  ^ O  and  BC.   (?) 

. '.  0  is  equally  distant  from  AB  and  BC. 

0  is  on  the  bisector  BF.     (§  231.)  q.e.d. 


PLANE   GEOMETRY 


Proposition  XXXIX.     Theorem 

238.  The  line  joining  the  middle  points  of  two  sides  of 
a  triangle  is  parallel  to  the  third  side,  and  equal  to  one 
half  of  it. 


Q.E.D. 


Let  DE  join  the  middle  points  of  AB  and  BC. 
To  Prove  LE  II  to  AC,  and  DE  =  \AC. 
Proof.     Prolong  DE  until  EF  =  DE.     Draw  FC. 
Prove  A  BDE  and  EEC  equal  in  all  respects. 
Whence  DB  =  FC  and  Z  3  =  Z  4. 

FC  =  AD.      (?)  FC  is  II  to  AD.      (?) 

ADFC  is  a  parallelogram.     (?) 
.*.  DE  is  II  to  ^C. 
Prove  that  DE  =  iAC. 

239.  Corollary  I.  If  a  line  is  drawn  through  the  middle 
point  of  one  side  of  a  triangle  parallel  to  the  base,  it  bisects  the 
other  side,  and  is  equal  to  one  half  the  base. 

Let  DE  be  drawn  from  the  middle  point  B 

of  ^C  II  to  AC. 

To    Prove    DE    bisects    AB,    and  DE 
=  iAC. 

Proof.     Draw  EF  II  to  AB. 

Prove  Adbe=Afec.  ^ 

Whence  EF  =  DB  and  DE  =  FC 

EF=:AD.      (?) 

D  is  the  middle  point  of  AB.   (?)     DE  =  iAC.   (?) 


Q.E.D. 


BOOK  I 


69 


K--'- 


.^-c 


E.-/-/ 


M       N 


240.    Corollary  II.    To  divide  a  line  into  any  oiumber  of 
equal  parts. 

Let  AB  be  the  given 
line. 

Required  to   divide  it 
into    any    number,    say  A- 
five,  equal  parts. 

Draw  AC,  making  any  convenient  angle  with  AS. 

On  AC  lay  off  five  equal  distances,  AD,  DE,  EF,  FG,  and  GH. 

Draw  HB. 

Draw  GS,  FR,  EN,  and  DM  parallel  to  HB. 

AB  is  divided  into  five  equal  parts. 

VioyqAM=MN{^  239). 

Draw  MT  II  to  AC. 

Prove  MN  =  NR  (§  239). 

In  a  similar  manner  prove  NR  =  RS,  and  RS  =  SB.        q.e.f. 


241.  Exercise.  The  lines  joining  the 
middle  points  of  the  three  sides  of  a  tri- 
angle, divide  it  into  four  triangles  equal  in 
all  respects. 

Prove     A1  =  A2  =  A3  =  A4. 


242.  Exercise.  Perpendiculars  drawn 
from  the  middle  points  of  tw^o  sides  of  a 
triangle  to  the  third  side  are  equal. 

Prove  DF  =  EG. 


243.    Exercise.     The  lines  joining  the    B 
middle  points  of  the  sides  of  a  quadrilateral 
form  a  parallelogram,  equal  in  area  to  one 
half  the  quadrilateral.  E 

Use  §  238  to  prove  EFGH  a  parallelo- 
gram. 

Use  §  241  to  prove  EFGH  =  |  ABCD. 


70  PLANE   GEOMETRY 

244.  Exercise.  The  medial  lines  of  a  triangle  intersect  in  a  common 
point. 

Draw  two  medial  lines  AE  and  CD.  B 

Prove    that   they    meet    (§    127)    in    some 
point  0. 

Draw  BO  and  prolong  it. 

It  is  required  to  show  that  F  is  the  middle  y'O--'^'^?^ 

point  of  ^C.  f^A 

Draw  AHW  to  DC,  and  prolong  BF  until  it  ""--.^  ^/ 

meets  AH.  h' 

Draw  HC. 

Prove  B0=  OH,  by  using  AABH. 

In  A  HBC,  prove  OJS:  parallel  to  HC. 

AOCHis  a  parallelogram.     .-.  JFis  the  middle  point  of  AC.         q.e.d. 

245.  Exercise.  The  point  of  intersection  of  the  medial  lines  divides 
each  median  into  two  segments  that  are  to  each  other  as  two  is  to  one. 

246.  Exercise.  Given  the  middle  points  of  the  sides  of  a  triangle,  to 
construct  the  triangle. 

As  the  variety  of  exercises  in  Geometry  is  practically  1111- 
limited,  it  is  impossible  to  give  for  their  solution  any  general 
rules,  as  may  usually  be  done  for  problems  in  Elementary 
Algebra  or  Arithmetic.  Yet  the  following  hints  may  be  of 
use  to  the  beginner : 

1.  Thoroughly  digest  all  the  facts  of  the  statement,  separat- 
ing clearly  the  hypothesis  from  the  conclusion. 

2.  Draw  a  diagram  expressing  all  of  these  facts,  including 
what  is  to  be  proved. 

3.  Draw  any  auxiliary  lines  that  may  seem  to  be  necessary 
in  the  proof.  ^ 

4.  Assuming  the  conclusion  to  be  true,  try  to  deduce  from 
it  simpler  relations  existing  between  the  parts  of  the  figure, 
and  finally  some  relation  that  can  be  established.  (This  is  the 
Analysis  of  the  Proposition.) 

iThe  student  should  remember  in  drawing  auxiliary  lines  that  a 
straight  line  may  be  drawn  fulfilling  only  tivo  conditions.  Two  condi- 
tions are  said  to  determine  a  straight  line. 


BOOK  I 


71 


5.  Then,  starting  with  the  relation  established,  reverse  the 
analysis,  tracing  it  back,  step  by  step,  until  the  conclusion  is 
reached. 


EXERCISES 

1.  If  two  angles  of  a  quadrilateral  are 
supplementary,  the  other  two  are  also 
supplementary. 

2.  Two  parallels  are  cut  by  a  trans- 
versal. Prove  that  the  bisectors  of  two 
interior  angles  on  the  same  side  are  per- 
pendicular to  each  other. 


3.  An  exterior  base  angle  of  an  isosceles  triangle 
is  1^  R.A.'s.     Find  the  angles  of  the  triangle. 

4.  If  the  angles  adjacent  to  one  base  of  a  trapezoid 
are  equal,  the  angles  adjacent  to  the  other  base  are 
also  equal.     [§122.] 


5.  In  the  parallelogram  ABCD,  AE  and 
CF  are  drawn  perpendicular  to  the  diagonal 
BB.     Prove  AE  =  CF. 


6.  ABC  and  CBD  are  two  supplementary 
adjacent  angles.  EB  bisects  ZABC,  and 
BF  is  perpendicular  to  EB.  Prove  that  BF 
bisects  Z  CBD. 

7.  Construct  a  right-angled  triangle,  hav- 
ing given  the  hypotenuse  and  one  of  the 
acute  angles. 

8.  Trisect  a  right  angle. 

9.  Construct  an  isosceles  triangle,  having 
given  the  base  and  the  vertical  angle. 

Suggestion.   Find  the  base  angles. 


72 


PLANE   GEOMETRY 


10.  ABC  is  an  isosceles  triangle,  and  BE 
is  parallel  to  AC.  Prove  that  BE  bisects 
the  exterior  angle  CBD. 


11.  The  lines  joining  the  middle  points  of 
the  opposite  sides  of  a  quadrilateral  bisect 
each  other.     [§  243.] 


12.  From  any  point  D  on  the  base  of  the 
isosceles  triangle  ABC,  BE  and  BE  are  drawn 
parallel  to  the  equal  sides  BC  and  AB  respec- 
tively. Prove  that  the  perimeter  of  DEBF  is 
constant  and  equals  AB  +  BC, 


13.  The  angle  formed  by  the  bisectors  of 
two  consecutive  angles  of  a  quadrilateral  is  equal 
to  one  half  the  sum  of  the  other  two  angles, 

[§§  138  and  157.] 

14.  How  many   sides  has   the    polygon    the    b 
sum  of  whose  interior  angles  exceeds  the  sum  of 
its  exterior  angles  by  12  right  angles  ? 

15.  On  the  sides  of  the  square  ABCD,  the 
equal  distances  AE,  BE,  CG,  and  DH  are  laid 
off.    Prove  that  the  quadrilateral  EFGH  is  also 

a  square.  A 


BOOK  I 


73 


16.  The  perpendiculars  erected  to  the  sides  of 
a  triangle  at  their  middle  points  meet  in  a  com- 
mon point. 

Suggestion.  Show  that  two  of  the  ±'s  meet. 
Then  show  that  the  third  ±  passes  through  their 
point  of  meeting.     [§  48.] 


17.  The  middle  point  of  the  hypotenuse 
of  a  right-angled  triangle  is  equally  distant 
from  the  three  vertices. 

Suggestion.  Draw  CD,  making  /.!=/.  A. 
Prove  Z 2  =  Z  ^,  and  ^D  =  DO  =  DB.  A 


18.  The  lines  joining  the  middle  points  of 
the  consecutive  sides  of  a  rhombus  form  a 
rectangle,  which  is  not  a  square. 

19.  From  two  points  on  the  same  side  of 
a  line  draw  two  lines  meeting  in  the  line  and 
making  equal  angles  with  it. 


20.  Prove  that  the  sum  ot  AC  and  BC 
(the  lines  that  make  equal  angles  with  xy)  is 
less  than  the  sum  of  any  other  pair  of  lines 
drawn  from  A  and  B  and  meeting  in  xy. 

Prolong  BC  until  CE  =  AC.  Prove  AD 
=  DE.     Then  apply  §  108  to  A  BDE. 


21.  If  the  base  of  an  isosceles  triangle  is 
prolonged,  twice  the  exterior  angle  =  2  R.A.'s 
+  the  vertical  angle  of  the  triangle. 

22.  In  the  triangle  ABC,  BD  is  drawn 
perpendicular  to  AC.  Prove  that  tlie  differ- 
ence between  Z  2  and  Z 1  equals  the  difference 
between  Z  A  and  Z  C. 


74 


PLANE   GEOMETRY 


23.  Given  the  sum  of  the  diagonal  and  a  side  of  a 
square,  construct  the  square. 

24.  If  BE  is  parallel  to  the  base  AG  oi  the  triangle 
ABC,  and  also  bisects  the  exterior  angle  CBD,  prove 
that  the  triangle  ABC  is  isosceles. 

25.  Given  the  difference  between  the  di- 
agonal and  a  side  of  a  square,  construct  the 
square. 


26.  Draw  DE  parallel  to  the  base  of  the 
triangle  ABC  so  that  DE  =  DA  +  EC. 

Two  constructions.  BE  may  cut  the  pro- 
longed sides. 

27.  ABCD  is  a  trapezoid.  Through  E, 
the  middle  of  CD,  draw  FG  parallel  to  BA 
and  meeting  BC  produced  at  F. 

Prove  the  parallelogram  ABFG  equal  in 
area  to  the  trapezoid  ABCD. 


28.  The  angle  formed  by  the  bisectors  of  two 
angles  of  an  equilateral  triangle  is  double  the  third 
angle. 

29.  In  the  isosceles  triangle  AB  C  draw  DE  paral- 
lel to  the  base  AC,  so  that  DA  =  DE  =  EC. 

30.  If  the  diagonals  of  a  parallelogram  are  equal 
and  perpendicular  to  each  other,  the  figure  is  a  square 


31.  If  from  a  point  on  the  base  of  an  isos- 
celes triangle  perpendiculars  are  drawn  to  the 
two  equal  sides,  their  sum  is  equal  to  a  per- 
pendicular drawn  from  either  extremity  of 
the  base  to  the  opposite  side. 

Suggestion.  Draw  PG  II  to  BC.  Prove 
^  AEP  and  AGP  equal. 


BOOK  I 


75 


32.  If  from  a  point  on  the  prolonged  base 
of  an  isosceles  triangle  perpendiculars  are 
drawn  to  the  two  equal  sides,  their  difference 
is  equal  to  a  perpendicular  drawn  from  either 
extremity  of  the  base  to  the  opposite  side. 


33.  In  the  triangle  ABC,  AE  and 
CE  are  the  bisectors  of  ZA  and  the 
exterior  angle  BCD  respectively. 

Prove  ZE=IZB. 


34.  If  one  angle  of  a  right-angled  triangle 
is  double  the  other,  the  hypotenuse  is  double 
the  shorter  leg. 

[See  Exercise  17.] 


35.    Construct  an  equilateral  triangle,  having  given  its  altitude. 

Br 


36.  The  quadrilateral  formed  by  the  bi- 
sectors of  the  angles  of  a  quadrilateral  has  its 
opposite  angles  supplementary. 

[See  Exercise  13.] 


37.    If  the  quadrilateral  ABCD  (see  figure  of  Ex.  36)  is  a  parallelo- 
gram, EFOH  is  a  rectangle. 


38.    If  the  quadrilateral  ABCD  (see  figure  of  Ex. 
EFGH  is  a  square. 


is  a  rectangle, 


39.    The  bisectors  of  the  exterior  angles  of  a  quadrilateral  form  a  second 
quadrilateral  whose  opposite  angles  are  supplementary. 


76 


PLANE    GEOMETRY 


40.  The  altitudes  of  a  triangle  meet 
in  a  common  point. 

Suggestion.  Through  the  three  ver- 
tices of  the  A  ABC  draw  parallels  to  the 
opposite  sides,  forming  A  GHL  Show 
that  the  altitudes  of  A  ABC  are  Js  to  the 
sides  of  A  GHI,  at  their  middle  points. 


:^'H 


41.  If  the  number  of  sides  of  an  equiangular  polygon  is  increased  by 
four,  each  angle  is  increased  by  ^  of  a  right  angle.  How  many  sides  has 
the  polygon?     [§158.] 


42.  In  the  parallelogram  ABCD,  BE  bi- 
sects AD  and  BF  bisects  BC.  Prove  that 
BE  and  DF  trisect  the  diagonal  AC. 

[§239.] 


43.  In  the  equilateral  triangle  ABC,  the 
distances  AD,  CF,  and  BE  are  equal.  Prove 
the  triangle  DEF  equilateral. 


44.  AF  and  HC  bisect  the  exterior  angles 
D AC  a,nd  ACE,  and  BG  bisects  the  interior 
angle  B  of  the  triangle  ABC.  Prove  that 
AF,  CH,  and  BG  meet  in  a  common  point. 

[See  §  233.] 


45.   If  two  lines  that  are  on  opposite  sides 
of  a  third  line  meet  at  a  point  of  that  third 
line,  making  the  non-adjacent  angles  equal,  ^ 
the  two  lines  form  one  and  the  same  line. 


BOOK  I 


77 


46.  What  is  the  greatest  number  of  acute  angles  a  convex  polygon  can 
have? 

Suggestion.     Show  that  if  there  were  more  than  three  acute  angles 
the  sum  of  the  exterior  angles  of  the  polygon  would  exceed  4  R.A.'s. 

47.  Given  two  lines  that  would  meet  if  sufficiently  produced,  draw  the 
bisector  of  their  angle,  without  prolonging  the  lines. 

D 


48.  Construct  a  triangle,  having  given  one 
angle,  one  of  its  including  sides,  and  the  sum 
of  the  other  two  sides. 


49.   Construct  a  triangle,  having  given  one  angle,  one  of  its  including 
sides,  and  the  difference  of  the  other  two  sides. 


The  side  opposite  the  given  angle  may  be  less  than  the  other  unknown 
side  (see  Fig.  1),  or  it  may  be  greater  than  the  other  unknown  side 
(see  Fig.  2).  g 


50.  BE  is  the  bisector  of  Z  ABC,  and  BD 
is  an  altitude  of  the  triangle  ABC.  Prove 
that  Z  1  is  one  half  the  difference  between 
the  base  angles  A  and  C. 


E  D 


51.  Through  a  point  draw  a  line  that  shall  be  equally  distant  from  two 
given  points.     [Two  ways.]  g 

52.  The  line  joining  the  middle  points  of 
two  opposite  sides  of  a  quadrilateral  bisects 
the  line  joining  the  middle  points  of  the 
diagonals. 

Suggestion.  Prove  that  EGFH  is  a  paral- 
lelogram. 


78 


PLANE   GEOMETRY 


63.    Of  all  triangles  having  the  same  base  and  equal  altitudes  the 
isosceles  triangle  has  the  least  perimeter.     [See  Ex.  20.] 


54.  Construct  a  triangle, 
having  given  the  perimeter 
and  the  two  base  angles. 


D-^ 


55.   Construct  a  triangle,  having  given  the  lengths  of  the  three  medians. 
[§§  244  and  245.] 


56.  If  the  diagonals  of  a  trapezoid  are 
equal,  the  non-parallel  sides  are  equal. 

BM  and  CN  are  each  A.  to  AD. 
Prove         A  A  CN=A  DBM, 
and  A  ABM  =  A  DON. 

57.  In  the  equilateral  triangle  ABC,  AD 
and  DC  bisect  the  angles  at  A  and  C.  DE 
is  drawn -11  to  AB,  and  DFW  to  BC.  Prove 
that  AC  \b  trisected. 


58.  AE  and  CD  are  perpendiculars  drawn 
from  the  extremities  of  ^C  to  the  bisector  of 
/.  B.  FD  and  FE  join  the  feet  of  these  per- 
pendiculars with  the  middle  point  of  AC. 

Prove    FD  =  FE  =  l{AB-BC). 


59.  ABC  is  a  R.A.  A,  AD  is  perpendicu- 
lar to  BC,  and  AE  is  the  median  to  BG. 
AF  bisects  angle  DAE. 

Prove  that  AF  also  bisects  angle  BAG. 


BOOK  II 


247.  Definitions.  A  circle  is  a  portion  of  a  plane  bounded 
by  a  curved  line,  all  the  points  of  which  are  equally  distant 
from  a  point  within  called  the  center. 

The  bounding  line  is  called  the  circum- 
ference. 

A  straight  line  from  the  center  to  any  point 
in  the  circumference  is  a  radius.  It  follows 
from  the  definition  of  circle  that  all  radii  of 
the  same  circle  are  equal. 

A  straight  line  passing  through  the  center  and  limited  by 
the  circumference  is  a  diameter. 

Every  diameter  is  composed  of  two  radii ;  therefore  all  diam- 
eters of  the  same  circle  are  equal. 

An  arc  is  any  portion  of  a  circumference. 

A  chord  is  a  straight  line  joining  the  extremities  of  an  arc. 

A  chord  is  said  to  subtend  the  arc  whose  extremities  it  joins, 
and  the  arc  is  said  to  be  subtended  by  the 
chord. 

Every  chord  subtends  two  different  arcs;  a> 
thus   the   chord  AB    subtends  the  arc  ANB, 
and  also  the  arc  AMB.     Unless  the  contrary 
is  specially  stated,  we  shall  assume  the  chord 
to  belong  to  the  smaller  arc. 

An  inscribed  polygon  is  a  polygon  whose 
vertices  are  in  the  circumference  and  whose 
sides  are  chords. 

[The  polygon  ABCD  is  inscnbed  m  the 
circle;  the  circle  is  also  said  to  be  circum- 
scribed about  the  polygon.] 

79 


80 


PLANE    GEOMETRY 


Proposition  I.     Problem 
248.  To  find  the  center  of  a  given  circle. 

D 


Let  xyz  be  the  given  circle. 

Required  to  find  its  center. 

Join  any  two  points  on  the  circumference,  as  A  and  B,  by 
the  line  AB. 

Bisect  AB  by  the  perpendicular  DC, 

Bisect  DC. 

Then  is  0  the  center  of  the  circle. 

By  definition,  the  center  of  the  circle  is  equally  distant 
from  A  and  B. 

By  §  48  the  center  is  on  DC. 

By  definition  the  center  of  the  circle  is  equally  distant  from 
D  and  C. 

Since  the  center  is  on  DC,  and  is  also  equally  distant  from 
D  and  C,  it  must  be  at  the  middle  point  of  DC,  that  is, 
at  0. 

Therefore,  0  is  the  center  of  the  circle  xyz.  q.e.f. 

249.  Corollary.  A  line  that  is  perpendicular  to  a  chord 
and  bisects  it,  passes  through  the  center  of  the  circle. 

Note.  It  follows  from  §  249  that  the  only  chords  in  a  circle  that  can 
bisect  each  other  are  diameters. 


BOOK   II  81 

250.  Exercise.     Describe  a  circumference  passing  through  two  given 
points. 

How  many  different  circumferences  can  be  described  passing  through 
two  given  points  ? 

251.  Exercise.     Describe  a  circumference,  with  a  given  radius,  and 
passing  through  two  given  points. 

How  many  circumferences  can  be  described  in  this  case  ? 
What  limit  is  there  to  the  length  of  the  given  radius  ? 

Proposition  II.     Theorem 

252.  A  diameter  divides  a  circle  and  also  its  circum- 
ference into  two  equal  parts. 


Let  AB  be  a  diameter  of  the  circle  whose  center  is  0. 

To  Prove  that  AB  divides  the  circle  and  also  its  circumfer- 
ence into  two  equal  parts. 

Proof.     Place  ACB  upon  ADB  so  that  AB  is  common. 

Then  will  the  curves  ACB  and  ADB  coincide,  for  if  they  do 
not  there  would  be  points  in  the  two  arcs  unequally  distant 
from  the  center,  which  contradicts  the  definition  of  circle. 

Therefore  AB  divides  the  circle  and  also  its  circumference 
into  two  equal  parts.  q.e.d. 

253.  Exercise.  Through  a  given  point  draw  a  line  bisecting  a  given 
circle. 

When  can  an  infinite  number  of  such  lines  be  drawn  ? 

SANDEKS'    GEOM. — 6 


82 


PLANE   GEOMETRY 


Propositiox  III.     Theorem 

254.  A  diameter  of  a  circle  is  greater  than  any  other 
chord. 


Let  AB  be  a  diameter  of  the  O  whose  center  is  O,  and  CD 
be  any  other  chord. 

To  Prove  AB  >  CD. 

Proof.     Draw  the  radii  OC  and  OD. 

Apply  §  168  to  A  OCD,  recollecting  that  AB  =  OC  +  OD. 

Q.E.D. 

255.  Exercise.     Prove   this  Proposition  (§  254),  using   a  figure  in 
which  the  given  chord  CD  intersects  the  diameter  AB. 

256.  Exercise.     Through  a  point  witliin  a  circle  draw  the  longest 
possible  chord. 

257.  Exercise.  The  side  AC  of  an  inscribed 
triangle  ABC  is  a  diameter  of  the  circle.  Compare 
the  angle  B  with  angles  A  and  C. 

258.  Exercise.  AB  is  perpendicular  to  the 
chord  CD.  and  bisects  it. 


Prove 


AB  >  CD. 


259.    Exercise.     The  diameter  AB  and    £ 
the  chord  CD  are  prolonged  until  they  meet 
at  ^. 

Prove  EA  <  EC 

and  EB  >  ED. 


BOOK  II  83 

Proposition  IV.     Theorem 

260.  A  straight  line  cannot  intersect  a  circumference 
in  more  than  two  points. 


Let  CDR  be  a  circumference  and  AB  d^  line  intersecting  it  at 
C  and  D. 

To  Prove  that  AB  cannot  intersect  the  circumference  at  any- 
other  point. 

Proof.  Suppose  that  AB  did  intersect  the  circumference  in 
a  third  point  E. 

Draw  the  radii  to  the  three  points. 

Now  we  have  three  equal  lines  (why  equal?)  drawn  from 
the  point  0  to  the  line  AB,  which  contradicts  (?). 

Therefore  the  supposition  that  AB  could  intersect  the  circum- 
ference in  more  than  two  points  is  false.  q.e.d. 

261.  Exercise,  Show  by  §§  249  and  92  that  AB  cannot  intersect  the 
circumference  in  three  points  (  O,  Z>,  and  E ) . 

A, 


262.  Definition.  A  secant  is  a  straight 
line  that  cuts  a  circumference. 


84  PLANE   GEOMETRY      , 

Proposition  Y.     Theorem 

263.  Circles  having  equal  radii  are  equal;  and  con- 
versely, equal  circles  have  equal  radii. 


Let  the  (D  whose  centers  are  0  and  G  have  equal  radii. 

To  Prove  the  ©  equal. 

Proof.  Place  the  O  whose  center  is  0  upon  the  O  whose 
center  is  C,  so  that  their  centers  coincide. 

Then  will  their  circumferences  also  coincide,  for  if  they  do 
not,  they  would  have  unequal  radii,  which  contradicts  the 
hypothesis. 

Since  the  circumferences  coincide  throughout,  the  circles  are 
equal.  q.e.d. 

Conversely.     Let  the  circles  be  equal. 
To  Prove  that  their  radii  are  equal. 

Proof.  Since  the  circles  are  equal,  they  can  be  made  to 
coincide. 

Therefore  their  radii  are  equal.  q.e.d. 

264.  Exercise.  Circles  having  equal  diameters  are  equal ;  and  con- 
versely, equal  circles  have  equal  diameters. 

265.  Exercise.  Two  circles  are  described  on  the  diagonals  of  a  rec- 
tangle as  diameters.     How  do  the  circles  compare  in  size  ? 

266.  Exercise.  If  the  circle  described  on  the  hypotenuse  of  a  right- 
angled  triangle  as  a  diameter  is  equal  to  the  circle  described  with  one  of 
the  legs  as  a  radius,  prove  that  one  of  the  acute  angles  of  the  triangle  is 
double  the  other. 


BOOK  II  85 


Proposition  VI.     Theorem 


267.  In  the  same  circle  or  in  equal  circles,  radii  form- 
ing equal  angles  at  the  center  intercept  equal  arcs  of 
the  circumference;  and  conversely,  radii  intercepting 
equal  arcs  of  the  circumference  form  equal  angles  at 
the  center. 


Let  ABC  and  BEF  be  two  equal  angles  at  the  centers  of  equal 
circles. 

To  Prove  arc  CA  =  arc  DF. 

Proof.  Place  the  circle  whose  center  is  B  upon  the  circle 
whose  center  is  E,  so  that  Z  B  shall  coincide  with  its  equal 
Ze. 

Since  the  radii  are  equal,  A  will  fall  upon  D  and  C  upon  F. 

The  arc  AC  will  coincide  with  the  arc  BF.     (Why  ?) 

Therefore  the  arc  AC  =  arc  DF.  q.e.d. 

Conversely.     Let     arc  CA  =  arc  DF. 

To  Prove  Zabc  =  Zdef. 

Proof.  Place  the  circle  whose  center  is  B  upon  the  circle 
whose  center  is  E,  so  that  the  circles  coincide,  and  the  arc  AC 
coincides  with  its  equal  arc  DF. 

BC  will  then  coincide  with  EF  (?)  and  AB  with  DE.     (?) 

Consequently  the  angles  ABC  and  DEF  coincide  and  are 
equal.  q.e.d. 

268.  Exercise.  Two  intersecting  diameters  divide  a  circumference 
into  four  arcs  which  are  equal,  two  and  two. 


86 


PLANE   GEOMETRY 


Proposition  VII.     Theorem 

269.  In  the  same  circle,  or  in  equal  circles,  if  tiuo  arcs 
are  equal,  the  chords  that  subtend  them  are  also  equal; 
and  conversely,  if  two  chords  are  equal,  the  arcs  that 
are  subtended  by  them  are  equal. 


Let  ABC  and  DEF  be  two  equal  arcs  in  the  equal  ©  whose 
centers  are  x  and  y. 

To  Prove  chord  AC  ==  chord  DF. 

Proof.   Draw  the  radii  xA,  xC,  yD,  and  yF, 
Show  that  Z1=Z2. 
Prove  A  AxC  and  DyF  equal. 

Whence 


AC  =  DF. 

Conversely.     Let  chord  AC  =  chord  I)F. 
To  Prove  arc  ABC  =  arc  DFF. 
Proof.     Draw  the  radii  xA,  xC,  yD,  and  yF. 
Prove  A  AxC  and  DyF  equal. 

Whence  Z1  =  Z2. 

.-.  arc  ABC  =  arc  DFF.     (?) 


q.e.d. 


q.e.d. 


270.  Exercise.  If  the  circumference  of  a  circle  is  divided  into  four 
equal  parts  and  their  extremities  are  joined  by  chords,  the  resulting 
quadrilateral  is  an  equilateral  parallelogram. 


BOOK  II  87 

Proposition  VIII.     Theorem 

271.  In  the  same  circle,  or  in  equal  circles,  if  two  arcs 
are  unequal  and  each  is  less  than  a  semi-circumfer- 
ence, the  greater  arc  is  subtended  by  the  greater  chord ; 
and  conversely,  the  greater  chord  subtends  the  greater 
arc. 


Let  M  and  N  be  the  centers  of  equal  circles  in  which  arc 

ABOdiVC  DEF. 

To  Prove  chord  J  C  >  chord  DF. 

Proof.     Draw  the  diameters  AG  and  DH. 

Place  the  semicircle  ACG  so  that  it  shall  coincide  with  the 
semicircle  DFH,  A  falling  on  D  and  G  on  H. 

Because  the  arc  ABC  is  greater  than  the  arc  DEF,  the  point 
C  will  fall  beyond  F  at  some  point  R,  the  chord  AC  taking  the 
position  DR. 

Draw  the  radii  NF  and  NR. 

Apply  §  181  to  A  DNF  and  DNR,  proving 

DR>DF.      .:  AC>  DF.  Q.E.D. 

Conversely.     Let  chord  AC>  chord  DF. 

To  Prove  arc  ^5C>  arc  DEF. 

Proof.  Show  that  the  arc  ABC  can  neither  be  equal  to  the 
arc  DEF  nor  less  than  it,  .-.  the  arc  ABC  must  be  greater  than 
the  arc  DEF.  q.e.d. 


88 


PLANE   GEOMETRY 


272.    Exercise.     ^jBC  is  a  scalene  triangle.    How 
do  the  arcs  AB^  BC,  and  AC  compare  ? 


273.  Exercise.  Give  a  direct 
proof  for  the  converse  of  Prop. 
VIII. 

[Draw  the  radii  and  show  that 
ZAMC  is  less  than  ZBND. 
Then  place  one  circle  upon  the 
other,  etc.] 


Propositio:^^  IX.     Theorem 

274.  A   diameter    that   is   perpendicular   to   a   chord 
bisects  the  chord  and  also  the  arc  subtended  by  it. 


0 

/ 

4\ 

\ 

\./            1 

2          y 

^^ 

B 

Let  AB  be  a  diameter  ±  to  CD. 

To  Prove  CE  =  ED  and  arc  CB  =  arc  BB. 

Proof.     Draw  the  radii  OG  and  OB. 
Prove  A  GOE  and  OEB  equal. 

Whence  CE  —  ED  and  Z.3  =  Z 4. 

Show  that  arc  GB  =  arc  BD. 


Q.E.D. 


BOOK  II  89 

275.  Corollary  I.  TJie  diameter  AB  also  bisects  the  arc 
CAD. 

276.  Corollary  II.  Prove  the  six  propositions  that  can  be 
formulated  from  the  following,  data,  using  any  two  for  the 
hypothesis  and  the  remaining  two  for  the  conclusion. 

A  line  that 

1.  Passes  through  the  center  of  the  O. 

2.  Bisects  the  chord. 

3.  Is  perpendicular  to  the  chord. 

4.  Bisects  the  arc. 

[Prop.  IX.  itself  is  one  of  the  six  proposi- 
tions, and  is   formed  by  using  1  and  3  as  hypothesis,  and 

2  and  4  as  conclusion ;  and  the  statement  of  §  249  uses  2  and 

3  for  its  hypothesis  and  1  for  its  conclusion.] 

277.  Corollary  III.     Bisect  a  given  arc. 

278.  Exercise.  What  is  the  locus  of  the  centers  of  parallel  ch'^rds 
in  a  circle  ? 

279.  Exercise.  Perpendiculars  erected  at  the 
middle  points  of  the  sides  of  a  quadrilateral  in- 
scribed in  a  circle  pass  through  a  common  point. 
Is  this  true  for  inscribed  polygons  of  more  than  four 
sides  ? 

280.  Exercise.  Through  a  given  point  in  a  circle  draw  a  chord  that 
shall  be  bisected  at  the  point. 

281.  Exercise.  If  the  line  joining  the  middle  points  of  two  chords 
in  a  circle  passes  through  the  center  of  the  circle,  prove  that  the  chords 
are  parallel. 

282.  Exercise.  The  chord  AB  divides  the  circumference  into  two 
arcs  ACB  and  ADB.  (See  figure  of  §  276.)  If  CD  is  drawn  connecting 
the  middle  points  of  these  arcs,  prove  that  it  is  perpendicular  to  AB  and 
bisects  it. 


90 


PLANE   GEOMETRY 
Proposition  X.     Theorem 


283.  In  the  same  circle  or  in  equal  circles  equal  chords 
are  equally  distant  from  the  center;  and  conversely, 
chords  that  are  equally  distant  from  the  center  are  equal. 


Let  AB  and  CD  be  equal  chords  in  the  equal  circles  whose 
centers  are  M  and  N. 

To  Prove  AB  and  CD  equally  distant  from  the  centers. 

Proof.     Draw  MR  and  NS  1.  to  AB  and  CD  respectively. 
MR  and  NS  measure  the  distance  of  the  chords  from  the 
centers.     (§  223.) 

Draw  the  radii  MB  and  ND. 
Prove  the  A  MRB  and  NSD  equal. 

Whence  MR  =  NS.  q.e.d. 

Conversely.     Let  AB  and  CD  be  equally  distant  from  the 

centers  (MR  —  NS). 

To  Prove  AB  =  CD. 

Proof.     Prove  A  MEB  and  NSD  equal. 

Whence  RB  =  SD. 

Therefore  AB  =  CD.     (?)  q.e.d. 

284.   Exercise.     What  is  the  locus  of  the  centers  of  equal  chords  in  a 
circle  ? 


BOOK   II  91 


285.  Exercise.  AB  and  CD  are  two  intersect- 
iug  chords,  and  they  make  equal  angles  with  the 
line  joining  their  point  of  intersection  with  the 
center  of  the  circle.  How  do  AB  and  CD  compare 
in  length  ? 

D 

286.  Exercise.  If  two  equal  chords  intersect  in  a  circle,  the  seg- 
ments of  one  chord  are  equal  respectively  to  those  of  the  other. 

287.  Exercise.  If  from  a  point  without  a  circle  two  secants  are 
drawn  terminating  in  the  concave  arc,  and  if  the  line  joining  the  center 
of  the  circle  with  the  given  point  bisects  the  angle  formed  by  the  secants, 
the  secants  are  equal. 

288.  Exercise.  If  two  chords  intersect  in  a  circle  and  a  segment  of 
one  of  them  is  equal  to  a  segment  of  the  other,  the  chords  are  equal. 

289.  Exercise.  The  line  joining  the  center  of  a  circle  with  the  point 
of  intersection  of  two  equal  chords,  bisects  the  angle  formed  by  the 
chords. 

290.  Exercise.  Through  a  given  point  of  a  chord  to  draw  another 
chord  equal  to  the  given  chord. 

\^Suggestion.  —  A^^\y  §  285.] 

291.  Exercise.  Through  a  given  point  in  a  circle  only  two  equal 
chords  can  be  drawn. 

For  what  point  in  the  circle  is  this  statement  untrue  ? 

292.  Exercise.  If  two  equal  chords  be  prolonged  until  they  meet 
at  a  point  without  the  circle,  the  secants  formed  are  equal. 

293.  Exercise.  Given  three  points  A^  B,  and  C  on  a  circumference, 
to  determine  a  fourth  point  X  on  that  circumference,  such,  that  if  the 
chords  AB  and  CX  be  prolonged  until  they  meet  at  a  point  without  the 
circle,  the  secants  formed  are  equal. 

294.  Exercise.  An  inscribed  quadrilateral  ABCD  has  its  sides  AB 
and  CD  parallel,  and  angles  D  and  C  equal. 

Prove  that  the  sides  AD  and  BC  are  equally  distant  from  the  center 
of  the  circle. 


92  PLANE   GEOMETRY 

Proposition  XI.     Theorem 

295.  In  the  same  circle  or  in  equal  circles,  the  smaller 
of  two  unequal  chords  is  at  the  greater  distance  from 
the  center ;  and  conversely,  of  two  unequal  chords,  the 
one  at  the  greater  distance  from  the  center  is  the  smaller. 


Let  M  and  N  be  the  centers  of  equal  (D,  and  let  AB  <  CD. 

To  Prove  that  AB  is  at  a  greater  distance  from  M  than  GB  is 
from  N. 

Proof.  Place  O  xAB  so  that  it  coincides  with  O  yCB^  B  fall- 
ing on  C  and  the  chord  AB  taking  the  position  CG. 

Draw  NS  and  NF  _L  to  GC  and  CD  respectively. 

Draw  8F. 

Prove  Z1>Z2. 

Whence  Z  3  <  Z  4.  (?) 

Whence  NS  >  I^F.  (?)  q.e.d. 

Conversely.     Let         JV5f  >  J^F. 
To  Prove  GC  <CD. 

Proof.  Z3<Z4.  (?) 

Z1>Z2.  (?) 

CF>SC.  (?) 

CD>  GC.  (?)  Q.E.D 

296.    Exercise.     Prove  the  converse  to  Prop.  XI.  indirectly. 
fShow  that  AB  can  neither  be  equal  to  nor  greater  than  CD.'] 


BOOK  II 


93 


297.  Exercise.     Through  a  point  within  a  circle  draw  the  smallest 
possible  chord. 

Proposition  XII.     Theorem 

298.  Through  three  points  not  in  the  same  straight 
line,  one  circumference,  and  only  one,  can  be  passed. 


(?) 
(?) 


Let  A,  B,  and  C  be  three  points  not  in  the  same  straight  line. 

To  Prove  that  a  circumference,  and  only  one,  can  be  passed 
through  A,  B,  and  C. 

Proof.     Draw  AB  and  BC. 

Bisect  AB  and  BChj  the  Js  DE  and  FG. 

Draw  DF. 

Show  that  Z  1  +  Z  2  <  2  E.A.'s. 

Whence  DE  and  FG  meet.     (?) 

0  is  equally  distant  from  A  and  B. 

0  is  equally  distant  from  B  and  C. 

Therefore  0  is  equally  distant  from  A,  B,  and  C. 

Therefore  a  circumference  described  with  0  as  a  center,  and 
with  OA,  OB,  or  OC  as  a  radius,  will  pass  thi*ough  A,  B,  and  C. 

The  line  DE  contains  all  the  points  that  are  equally  distant 
from  A  and  B.     (?) 

The  line  GF  contains  all  the  points  that  are  equally  distant 
from  B  and  C.     (?) 

Therefore  their  point  of  intersection  is  the  only  point  that  is 
equally  distant  from  A,  B,  and  C. 

Therefore  only  one  circumference  can  be  passed  through  A, 
By  and  C.  Q.E.D. 


94  PLANE    GEOMETRY 

299.  Corollary.  Two  circumferences  can  intersect  in  only 
two  points. 

300.  Exercise.  Why  cannot  a  circumference  be  passed  through  three 
points  that  are  in  a  straight  line  ? 

301.  Exercise.     Circumscribe  a  circle  about  a  given  triangle. 

302.  Exercise.  Show,  by  using  §§  298  and  249,  that  the  perpendicu- 
lars erected  to  the  sides  of  a  triangle  at  their  middle  points  pass  through 
a  common  point. 

303.  Exercise.     Find  the  center  of  a  given  circle  by  using  §  298. 

304.  Exercise.  From  a  given  point  without  a  circle  only  two  equal 
secants,  terminating  in  the  circumference,  can  be  drawn. 

Suggestion. — Suppose  that  three  equal  secants  could  be  drawn. 
Using  the  given  point  as  a  center  and  the  length  of  the  secant  as  a  radius, 
describe  a  circle.     Apply  §  299. 

305.  Exercise.  Circumscribe  a  circle  about  a  right-angled  triangle. 
Show  that  the  center  of  the  circle  lies  on  the  hypotenuse. 

306.  Definitions.  A  straight  line  is 
tangent  to  a  circle  when  it  touches  the  cir- 
cumference at  one  point  only.  The  point  at 
which  the  straight  line  meets  the  circum- 
ference is  called  the  point  of  tangency.  All 
other  points  of  the  straight  line  lie  without 
the  circumference.  ^The  circle  is  also  said 
to  be  tangent  to  the  line. 

Two  circles  are  tangent  to  each  other  when  their  circumfer- 
ences touch  at  one  point  only.     If  one  circle  lies  outside  of  the 


other,  they  are  tangent  externally;  if  one  circle  is  within  the 
other,  they  are  tangent  internally. 


BOOK  II  95 

Proposition  XIII.     Theorem 

307.  If  a  line  is  perpendicular  to  a  radius  at  its 
outer  extremity  it  is  tangent  to  the  circle  at  that  point; 
and  conversely,  a  tangent  to  a  circle  is  perpendicular 
to  the  radius  drawn  to  the  point  of  tangency. 


Let  ab\>q  1.  to  the  radius  CD  at  Z). 
To  Prove  AB  tangent  to  the  circle. 
Proof.     Connect  C  with  any  other  point  oi  AB  as  E. 
CE  >  CD.      (?) 

Since  CE  is  longer  than  a  radius,  E  lies  without  the 
circumference. 

E  is  any  point  on  AB  (except  D). 

Therefore  every  point  on  AB  (except  7))  lies  without  the  cir- 
cumference, and  AB  touches  the  circumference  at  D  only. 

Q.E.D. 

Conversely.     Let  AB  be  tangent  to  the  O  at  D. 

To  Prove  AB  JL  to  CD. 

Proof.     Connect  C  with  any  other  point  of  AB  as  E. 

Since  AB  is  tangent  to  the  circle  at  D,  E  lies  without  the 

circumference. 

CE>CD.      (?) 

CE  is  the  distance  from  C  to  any  point  of  AB  (except  D). 

CD  is  therefore  the  shortest  distance  from  C  to  AB. 

:.  CD  is  perpendicular  to  AB.  q.e.d. 


96  PLANE    GEOMETRY 

CoROLLAKY  I.  At  a  giveu  point  on  a  circumference  draw  a 
tangent  to  the  circle. 

Corollary  II.  At  a  point  on  a  circumference  only  one  tan- 
gent can  be  drawn  to  the  circle. 

308.  Exercise.  A  perpendicular  erected  to  a  tangent  at  the  point 
of  tangency  will  pass  through  the  center  of  the  circle. 

309.  Exercise.  If  two  tangents  are  drawn  to  a  circle  at  the  ex- 
tremities of  a  diameter,  they  are  parallel. 

310.  Exercise.  The  line  joining  the  points  of  tangency  of  two  par- 
allel tangents  passes  through  the  center  of  the  circle. 

311.  Exercise.  If  two  unequal  circles  have  the  same  center,  a  line 
that  is  tangent  to  the  inner  circle,  and  is  a  chord  of  the  outer,  is  bisected 
at  the  point  of  tangency. 

312.  Exercise.  Draw  a  line  tangent  to  a  circle  and  parallel  to  a 
given  line. 

313.  Exercise.  Draw  a  line  tangent  to  a  circle  and  perpendicular  to 
a  given  line. 

314.  Exercise.  If  an  equilateral  polygon  is  inscribed  in  a  circle, 
prove  that  a  second  circle  can  be  inscribed  in  the  polygon. 

315.  Exercise.  Circumscribe  about  a  given  circle  a  triangle  whose 
sides  are  parallel  to  the  sides  of  a  given  triangle. 

316.  Exercise.  To  construct  a  triangle  having  given  two  sides  and 
an  angle  opposite  one  of  them. 

Let  m  and  n  be  the  two  given  sides,  and  Z  s  the  angle  oppo- 
site side  n.  ^ 

Required  to  construct  the  A. 

Lay  off  an  indefinite  line 
AD.  At  A  construct  Za=Zs. 
Make  AB  =  m.  With  B  as 
a  center,  and  n  as  a  radius, 
describe  an  arc  intersecting 
AD  at  C.  Draw  BC.  Show 
that  A  ^J5C  is  the  required  A.  A-  ^^^ 


BOOK   II 


97 


Scholium.  When  the  given  angle 
is  acute,  and  the  side  opposite  the 
given  angle  is  less  than  the  perpen- 
dicular from  B  to  AD  J  there  is  no 
construction. 

When  the  given  angle  is  acute,  and 
the  side  opposite  the  given  angle  is 
equal  to  the  perpendicular  from  B  to 
AD,  there  is  one  construction,  and  the 
A  is  right-angled. 

When  the  given  angle  is  acute,  and 
the  side  opposite  the  given  angle  is 
greater  than  the  perpendicular  from 
B  to  AD  and  is  less  than  AB,  there  are 
two  constructions. 

Both  A  ABC  and  A  AB&  fulfill  the 
required  conditions. 

When  the  given  angle  is  acute,  and 
the  side  opposite  the  given  angle  is 
equal  to  AB,  there  is  one  construction. 

When  the  given  angle  is  acute,  and 
the  side  opposite  the  given  angle  is 
greater  than  AB,  there  is  one 
construction. 

A  ABC  fulfills  the  required 
conditions,  but  A  ABC'  does  , 
not. 

If  the  given  angle  is  ob- 
tuse, the  opposite  side  must  "~~~  —  '"" 
be  greater  than  AB  (?),  and  there  never  can  Lg  more  than  one 
construction. 


yo 


317.   Exercise.     Construct  a  triangle  ABC  in  which  AB  =  5  inches, 
ZA  =  ^  BA,  and  side  BC  =  1,  2,  3,  4,  and  5  inches  in  turn. 
State  the  number  of  solutions  in  each  case. 
How  long  must  ^C  be  in  order  to  form  a  right-angled  triangle  ? 

SANDERS'    GEOM. 7 


98 


PLANE    GEOMETRY 


Proposition  XIV.     Theorem 

318.  Parallel  lines  intercept  equal  arcs  of  a  circum- 
ference; and  conversely,  lines  intercepting  equal  arcs 
of  a  circumference  are  parallel. 


I.    Let  AB  and  CD  be  parallel  chords. 
To  Prove  arc  ^C  =  arc  BD. 

Proof.     Draw  the  diameter  EF  A.  to  AB. 
EF  is  A- to  CD.     (?) 
EA  =  EB  and  EC  =  ED.     (?) 
Whence  AC  =  BD. 

Conversely.     Let  AC  =  BD. 

To  Prove  AB  and  CD  parallel. 

Draw  the  diameter  EF  ±  to  AB. 

AE  =  EB.      (?) 

AC  =  BD.      (?) 

EC  =  ED.      (?) 

EF  is  ±  to  CD.      (?) 

AB  and  CD  are  parallel.     (?) 


q.e.d. 


q.e.d 


99 


F  F 

II.  Let  the  tangent  AB  and  tlie  chord  CD  be  parallel. 
To  Prove  CE  =  ED. 

Proof.     Draw  the  diameter  FE  to  the  point  of  tangency  E. 
FE  is  ±  to  AB.      (?) 
FE  is  ±  to  Ci>.     (?) 

CE  =  ED.  (?)  Q.E.D. 

Conversely.     Let  CE  =  ED. 

To  Prove  AB  and  CD  parallel. 

Proof.     Draw  the  diameter  FE  to  the  point  of  tangency  E. 

Prove  AB  and  CD  each  ±  to  EF.  q.e.d. 

III.  Let  the  tangents  AB  and  CD  be  parallel. 
To  Prove  EMF  =  ENF. 
Proof. 
Draw  the  chord  AT  II  to  AB. 

XT  is  11  to  CD.      (?) 
EX  =  EY  and  XF  =  YJ?^.      (?) 
EMF  =  ^JV^F.        Q.E.D. 
Conversely. 
Let      EMF  =  ENF. 

To  Prove  the  tangents  AB 
and  CD  parallel.  [The  proof 
is  left  to  the  student.] 


100 


PLANE    GEOMETRY 


319.  Exercise.  ABCD  is  a  trapezoid  inscribed  in  the  circle  whose 
center  is  0. 

Prove  tliat  the  non-parallel  sides  AB  and  CD 
are  equal. 

320.  Exercise.     Prove  the  converse  of  the  pre-      ^ 
ceding  exercise,  i.e.  if  tw^o  opposite  sides  of  an  in- 
scribed quadrilateral  are  equal,  the  quadrilateral  is  a 
trapezoid. 

321.  Exercise.  The  diagonals  of  an  inscribed 
trapezoid  are  equal. 

322.  Exercise.  The  side  AB  of  the  inscribed 
angle  ABC  is  a  diameter.  Prove  that  the  diameter 
DE  drawn  parallel  to  BC  bisects  the  arc  AC. 

Proposition  XY.     Theorem 

323.  If  two  circumferences  intersect  each  other,  the 
line  Joining  their  centers  bisects  at  right  angles  their 
common  chord. 


Let  AB  be  the  line  joining  the  centers  of  two  circumferences 
intersecting  at  C  and  D.  ^^— --^c 

To  Prove  AB  bisects  CB  at  right  angles. 
Proof.     Use  §  49. 

324-    Exercise.     Prove  §  32.3,  using  this  figure. 

325.   Exercise.     The   centers  of  all   circles  that 
pass  through  C  and  D  (figure  of  §  323)  are  on  AB  or  its  prolongation. 


BOOK   II  101 

Proposition  XVI.     Theorem 

326.  //  two  circles  are  tangent,  either  eocternally  or 
internally,  their  centers  and  the  point  of  tangency  are 
in  the  same  straight  line. 


Let  A  and  B  be  the  centers  of  two  (D  tangent  externally  at  C. 
To  Prove  that  A,  C,  and  B  are  in  the  same  straight  line. 
Proof.     Draw  the  radii  AC  and  BC  to  the  point  of  tangency. 
It  is  required  to  prove  that  ACB  is  a  straight  line. 
If  it  can  be  shown  that  A  CB  is  shorter  than  any  other  line 
joining  A  and  B,  then,  by  Axiom  14,  ACB  is  a  straight  line. 

I.  To  show  that  ACB  is  shorter  than  any  other  line  joining 
A  and  B  and  passing  through  C. 

Let  AinnB  be  any  other  line  joining  A  and  B  and  passing 
through  C.  AC-\-CB  <  AmC  +  CnB.      (?) 

or  ACB  <  AmnB. 

II.  To  show   that  ACB  is  shorter   than   any  line  joining 
A  and  B  and  not  passing  through  C. 

Join  A  and  B  by  any  line  ADB  not  passing  through  C. 

Since  the  circles  touch  at  C  only,  any  line  joining  the  centers 
and  not  passing  through  C  must  pass  outside  of  the  circles, 
and  must  be  greater  than  the  sum  of  the  radii. 
.-.    ACB  <  ADB. 

ACB  is  the  shortest  distance  between  A  and  B. 

.'.  ACB  \?>  ^  straight  line.  q.e.d. 


102 


PLANE    GEOMETRY 


Let  A  and  B  be  the  centers  of  two  circles  tangent  inter- 
nally at  C. 

To  Prove  that  A,  B,  and  C  are  in  a 
straight  line. 

Proof.  At  C  draw  DE  tangent  to  the 
outer  circle.     (?) 

All  the  points  of  DE  except  C  lie 
entirely  without  the  outer  circle,  and 
consequently  entirely  without  the  inner 
circle. 

DE  touches  the  inner  circle  at  C  only,  and  is  tangent  to  it 
also. 

Draw  the  radii  A  C  and  ^  C  to  the  point  of  tangency . 

AC  and  BC  are  each  ±  to  DE.     (?) 

A,  B,  and  C  are  in  a  straight  line.     (?)  q.e.d. 


327.  Corollary.  If  two  circles  are  tangent,  either  exter- 
nally or  internally,  and  if  at  their  point  of  tangency  a  line  is 
drawn  tangeyit  to  one  of  the  circles,  it  is  tangent  to  the  other 
also. 

328.  Exercise.  Two  circles  are  tangent,  and  the  distance  between 
their  centers  is  10  in.  The  radius  of  one  circle  is  4  in.  What  is  the 
radius  of  the  other  ?     (Two  solutions.) 


329.  Exercise.  Draw  a  common  tangent  to  two  circles  tangent 
to  each  other.     (§327.) 

How  many  common  tangents  can  be  drawn  to  two  circles  that  are 
tangent  internally  ?  Tangent  externally  ?  [In  the  latter  case  the  student 
is  expected  at  present  to  draw  only  one  of  the  three  common  tangents.] 


BOOK   II 


103 


Propositiox  XVII.     Theorem 

330.  a.  If  two  circles  are  entirely  without  each  other 
and  are  not  tangent,  the  distance  between  their  centers 
is  greater  than  the  sum  of  their  radii. 

h.  If  two  circles  are  tangent  externally,  the  distance 
between  their  centers  is  equal  to  the  sum  of  their  radii. 

c.  If  two  circles  intersect,  the  distance  between  their 
centers  is  less  than  the  sum  and  greater  than  the  dif- 
ference of  their  radii. 

d.  If  two  circles  are  tangent  internally,  the  distance 
between  their  centers  is  equal  to  the  difference  of  their 
radii. 

e.  If  one  circle  lies  wholly  within  another,  and  is  not 
tangent  to  it,  the  distance  between  their  centers  is  less 
than  the  difference  of  their  radii. 


AB  >  sum  of  radii.     (?) 


AB  passes  through  C.     (?) 
AB  =  sum  of  radii.         (?) 


Draw  the  radii  AC  and  BC. 
AB  <  sum  of  radii.  (?) 

AB  >  difference  of  radii.     (?) 


104 


PLANE   GEOMETRY 


AB  prolonged  passes  through  C.     (?) 
AB  =  difference  of  radii.      (?) 


AD  is  the  radius  of  the  large  O. 
-BC7  is  the  radius  of  the  small  O. 
What  is  tlie  difference  of  the  radii  ? 
AB  <C  difference  of  radii.     (?) 


[If  two  circles  are  concentric  (i.e.  have  the  same  center)  the 
distance  between  their  centers  is,  of  course,  zero.  This  position 
manifestly  comes  under  Case  e.'\ 

331.  Corollary.  State  and  prove  the  coiiverse  of  each  case 
of  Prop.  XVII.     [Indirect  proof.] 

332.  Exercise.  If  the  centers  of  two  circles  are  on  a  certain  Hne, 
and  their  circumferences  pass  through  a  point  of  that  line,  the  circles  are 
tangent  to  each  other. 

333.  Exercise.  Two  circles  whose  radii  are  6  in.  and  8  in.  respec- 
tively, intersect.  Between  what  limits  does  the  length  of  the  line  joining 
their  centers  lie  ? 

334.  Exercise.  With  a  given  radius  describe  a  circle  tangent  to  a 
given  circle  at  a  given  point.     [Two  solutions.] 

335.  Exercise.  What  is  the  locus  of  the  centers  of  circles  having  a 
given  radius  and  tangent  to  a  given  circle  ? 

336.  Exercise.  Describe  a  circle  having  a  given  radius  and  tangent 
to  two  given  circles. 

Draw  the  figures  for  the  next  three  constructions  accurately 
and  to  scale.     [1  ft.  =  |  in.] 

337.  ExERCisK.  A  and  B  are  the  centers  of  two  circles.  AB  =  7  ft., 
radius  of  O  ^  =  2  ft.,  and  radius  of  O  ^  =  3  ft.  Describe  a  circle,  with 
radius  2^  ft.,  tangent  to  both. 


BOOK  II  105 

338.  Exercise.  ^1  and  B  are  the  centers  of  two  circles.  AB  =  1\  ft., 
radius  of  O  ^  =  5  ft.,  and  radius  of  O  jB  =  2\  ft.  Describe  a  circle,  with 
radius  1|  ft.,  tangent  to  both. 

339-  Exercise.  Describe  three  circles,  with  radii  1  ft. ,  2  ft. ,  and  3  ft. 
respectively,  and  each  tangent  externally  to  both  of  the  others. 

340.  Definition.  The  ratio  of  one  quantity  to  another  of 
the  same  kind  is  the  quotient  obtained  by  dividing  the  numer- 
ical measure  of  the  first  by  the  numerical  measure  of  the 
second. 

The  ratio  of  5  ft.  to  7  ft.  is  f .  The  ratio  of  7  lb.  to  4  lb.  is 
-J,  or  IJ.  The  ratio  of  the  diagonal  of  a  square  to  a  side  is  V2 
(as  will  be  shown). 

It  is  necessary  that  the  two  quantities  be  of  the  same  kind; 
thus,  it  is  impossible  to  express  the  ratio  of  5  ft.  to  7  lb. 

Definitions.  A  constant  is  a  quantity  whose  value  remains 
unchanged  throughout  the  same  discussion. 

A  variable  is  a  quantity  whose  value  may  undergo  an  indefi- 
nite number  of  successive  changes  in  the  same  discussion. 

The  limit  of  a  variable  is  a  constant,  from  which  the  variable 
may  be  made  to  differ  by  less  than  any  assignable  quantity, 
but  which  it  can  never  equal. 

Suppose  a  point  to  move    a  C  D      E      B 

from  A  toward  B,  under  tlie 

condition  that  in  the  first  unit  of  time  it  shall  pass  over  one 
half  the  distance  from  A  to  B  ]  and  in  the  next  equal  unit  of 
time,  one  half  of  the  remaining  distance ;  and  in  each  succes- 
sive equal  unit  of  time,  one  half  the  remaining  distance. 

It  is  plain  that  the  point  would  never  reach  B,  as  there 
would  always  remain  half  of  some  distance  to  be  covered. 

The  distance  from  A  to  the  moving  point  is  a  variable,  which 
is  approaching  the  constant  distance  ^B  as  a  limit.  The  dif- 
ference between  the  variable  distance  and  the  constant  distance 
AB  can  be  made  less  than  any  assignable  quantity,  but  never 
can  be  made  equal  to  zero. 


106  PLANE   GEOMETRY 

Proposition  XVIII.     Theorem 

341.  If  two  variables  are  always  equal,  and  are  each 
approa^^hing  a  limit,  their  limits  are  equal. 

A  M  E  B 

I  I  II 


Let  AM  and  CN  be  two  variables  that  are  always  equal,  and 
let  AB  and  CD  be  their  respective  limits. 

To  Prove  AB  =  CD. 

Proof.     Suppose  AB  and  CD  to  be  unequal,  and  AB  >  CD. 

Lay  off  AE  =  CD. 

Now,  by  the  definition  of  limit,  AM  can  be  made  to  differ 
from  AB  by  less  than  any  assignable  quantity,  and  therefore 
by  less  than  EB. 

So  ^i^/  may  be  greater  than  AE. 

By  the  definition  of  limit,  CN  <  CD.  But  since  AE  =  CD, 
CN  <AE. 

Now  AM >  AE  and  CN  <  AE-,  but  by  hypothesis  AM  and  CN 
are  always  equal. 

The  result  being  absurd,  the  supposition  that  AB  and  CD  are 
unequal  is  false. 

Therefore  AB  and  CD  are  equal.  q.e.d. 

342.  Definition.  Two  magnitudes  are  commensurable  when 
they  have  a  common  unit  of  measure ;  i.e.  when  they  each  con- 
tain a  third  magnitude  a  whole  number  of  times. 

Two  magnitudes  are  incommensurable  when  they  have  no 
common  unit  of  measure ;  i.e.  when  there  ex- 
ists no  third  magnitude,  however  small,  that 
is  contained  in  each  a  whole  number  of  times. 


A 


343.  Definition.  A  sector  is  that  part  of 
a  circle  included  between  two  radii  and  their 
intercepted  arc. 


BOOK  II 


107 


Proposition  XIX.     Theorem 

344.    In  the  saiive  circle  or  in  equal  circles,  two  angles 
at  the  center  have  tiis  same  ratio  as  their  intercepted  arcs. 


Case  I 

When  the  angles  are  commensurable. 

Let  ABC  and  DBF  be  commensurable  angles  at  the  centers 
of  equal  (D. 

To  Prove 


Zabc 


DF 


Abef 

Proof.     Since  A  ABC  and  BEF  are  commensurable,  they  have 
a  common  unit  of  measure. 

Let  Z  X  he  this  unit,  and  suppose  it  is  contained  in  /.ABC 
m  times,  and  in  Z  BEF  n  times. 

Zabc     m^ 
n 


Whence 


(1) 


Zbef 

The  small  angles  into  which  A  ABC  and  BEF  are  divided 
are  equal,  since  each  equals  Z  x. 

By  §  267,  the  arcs  into  which  AC  and  BF  are  divided  by  the 
radii  are  equal. 

Since  AC  is  composed  of  m  of  these  equal  arcs,  and  BF  of  n 
of  these  equal  arcs,  j^c     m 

BF      n 
Apply  Axiom  1  to  (1)  and  (2). 

Zabc  _Ac 

Z  BEF  ~  BF 


(2) 


Q.E.D 


108  PLANE   GEOMETRY 


F 

Case  II 
When  the  angles  are  incommensurable. 

Let  ABC  and  DEF  be  two  incommensurable  angles  at  the 
centers  of  equal  ©. 

m    T^  /.ABC       AC 

To  Prove  — = 

Z.DEF       DF 

Proof.  Let  Z.  DEF  be  divided  into  a  number  of  equal  angles, 
and  let  one  of  these  be  applied  to  Z  ABC  as  a  unit  of  measure. 

Since  A  ABC  and  DEF  are  incommensurable,  ABC  will  not 
contain  this  unit  of  measure  exactly,  but  a  certain  number 
of  these  angles  will  extend  as  far  as,  say,  ABG,  leaving  a 
remainder  Z  GBC,  smaller  than  the  unit  of  measure. 

Since  A  ABG  and  DEF  are  commensurable,     (?) 

^^^  =  ^  by  Case  I. 

A DEF       DF 

By  increasing  indefinitely  the  number  of  parts  into  which 

A  DEF  is  divided,  the  parts  will  become  smaller  and  smaller, 

and  the  remainder  A  GBC  will  also  diminish  indefinitely. 

Now is  evidently  a  variable,  as  is  also  — ,  and  these 

AdEF  •"  DF 

variables  are  always  equal  to  each  other.     (Case  I.) 

The  limit  of  the  variable  —, is  — -' 

A DEF      A DEF 

The  limit  of  the  variable  —  is 

DF       DF 

By  §341,  ^ABCl^AC,  ^.^.^, 

''  A  DEF      DF 


BOOK  II  109 

345.  Corollary.  In  the  same  circle,  or  in  equal  circles, 
sectors  are  to  each  other  as  their  arcs.  [The  proof  is  analogous 
to  that  of  the  Proposition,  substituting  sector  for  angle.^ 

346.  Scholium.  If  two  diameters  are 
drawn  perpendicular  to  each  other,  four  right 
angles  are  formed  at  the  center  of  the  circle. 
By  §  267,  the  circumference  is  divided  into 
four  equal  arcs  called  quadrants. 

If  one  of  these  right  angles  were  divided 
into  any  number  of  equal  parts,  it  could 
be  shown  by  §  267,  that  the  quadrant  subtending  the  right 
angle  is  also  divided  into  the  same  number  of  equal  parts.  If, 
for  example,  the  right  angle  at  the  center  were  divided  into 
four  equal  parts,  the  arcs  intercepted  by  the  sides  of  these 
angles  would  each  be  one  fourth  of  a  quadrant ;  and  conversely, 
radii  intercepting  an  arc  that  is  one  fourth  of  a  quadrant,  form 
an  angle  at  the  center  which  is  one  fourth  of  a  right  angle. 

If  any  angle  as  Z  Z)  OM  be  taken  at  random  and  compared 
with  a  right  angle, 

B     S  S44-  Z.DOM _       DM 

E.  A.       quadrant' 
i.e.  the  angle  DOM  is  the  same  part  of  a  right  angle  that  its 
intercepted  arc  is  of  a  quadrant. 

In  this  sense  an  angle  at  the  center  is  said  to  be  measured  by 
its  intercepted  arc. 

347.  Scholium.  A  quadrant  is  usually  conceived  to  be 
divided  into  ninety  equal  parts,  each  part  called  a  degree  of  arc. 

The  angle  at  the  center  that  is  measured  by  a  degree  of  arc 
is  called  a  degree  of  angle. 

The  degree  is  divided  into  sixty  equal  parts  called  minides, 
and  each  minute  is  again  subdivided  into  sixty  equal  parts 
called  seconds. 

Degrees,  minutes,  and  seconds  are  designated  by  the  symbols 
°,  ',  "  respectively.  Thus,  49  degrees,  27  minutes,  and  35  sec- 
onds, is  written  49°  27'  35". 


110 


PLANE   GEOMETRY 


348.  Exercise.     Add  23°  46'  27"  and  19°  21'  36". 

349.  Exercise.     Subtract  15°  42'  39"  from  93°  16'  25". 

350.  Exercise.     How  many  degrees  in  an  angle  of  an  equilateral 
triangle  ? 

351.  Exercise.     Multiply  13°  27'  35"  by  3,  and  add  the  product  to 
one  half  of  12°  15'  10". 

352.  Exercise.     How  many  degrees  are  there  in  each  angle  of  an 
isosceles  right-angled  triangle  ? 

353.  Exercise.     Express  in  degrees,  minutes,  and  seconds  the  value 
of  one  angle  of  a  regular  heptagon. 


354.  Definition.  An  iyiscrihed  angle  is  an 
angle  whose  vertex  is  in  the  circumference 
and  whose  sides  are  chords. 


The  symbol  '^  is  used  for  the  phrase  is 
measured  by.  Thus,  Z  ABC  ~  arc  ^C  is  read : 
The  angle  ABC  is  measured  by  the  arc  AG. 


A  segment  is  that  part  of  a  circle  which  is 
include  between  an  arc  and  its  chord. 
[ACB  and  ADB  are  both  segments.] 

An  angle  is  inscribed  in  a  segment  when  its 
vertex  is  in  the  arc  of  the  segment  and  its 
sides  terminate  in  the  extremities  of  that  arc. 

[Z  ABC  and  Z.ADC  are  inscribed  in  the  seg- 
ment AmC.'] 


BOOK  II 


111 


Proposition  XX.     Theorem 

355.    An  inscribed  angle  is  measured  by  one  half  of 
the  arc  intercepted  by  its  sides. 


Let  Z.ABC  be  an  inscribed  angle  having  a  diameter  for  one 
of  its  sides. 

To  Prove  Z.ABC  <^\AC. 

Proof.     Draw  the  radius  0(7. 

Prove  A1  =  2Zb. 

Zl'^AC.     (§346.) 

.-.  Z  J5,  which  is  one  half  Z  1,  is  measured  by  one  half  the 


arc  A  C. 

Case  II 

Let  A  ABC  be  an  inscribed  angle  hav- 
ing the  center  between  its  sides. 

To  Prove    Zabc  ^^AC, 

Draw  the  diameter  BD. 

Z  ABD  ^  1  AB.     (Case  I.) 
ZbbC'^^dc.     (Case  I.) 

Z  ABC,  which  is  the  sum  of  A  ABD 
and  DBC,  is  measured  by  the  sum  of 
their  measures  (^  AD  -^^DC),  that  is, 
hj^AC,  q.e.d. 


Q.E.D. 


112 


PLANE   GEOMETRY 


Case  III 
Let  Z.ABC  be  an  inscribed  angle  hav- 
ing the  center  without  its  sides. 

To  Prove    Zabc  ^  }  A  C. 
Proof.     Draw  the  diameter  BD. 
ZdBC  ^^DC.  (?) 
/.DBA  '^^DA.  (?) 
/.ABC,  which  is   the  difference    be- 
tween  A  DBC  and  DBA,  is   measured 
by   the    difference    of    their    measures 
{^DC  —  ^DA),  that  is,  by  i  ^C.      q.e.d. 


356.    Corollary  I.     An 
the  same  segment  are  equal. 


les  inscribed 


357.  Corollary  II.  Angles  inscribed 
in  a  semicircle  are  right  angles. 

[/I'^^ABC.  But  1  of  the  arc  ABC  is 
a  quadrant.  Therefore,  by  §  346,  Z  1  is 
a  right  angle.] 


358.  Corollary  III.  A71  angle  inscribed 
in  a  segment  that  is  greater  than  a  semicircle 
is  acute. 


359.  Corollary  IY.  An  angle  inscribed 
in  a  segment  that  is  less  than  a  semicircle 
is  obtuse. 


BOOK   II  113 

360.  Corollary  V.  The  opjyosite  angles 
of  an  inscribed  quadrilateral  are  supple- 
mentary. 

[Show  that  the  sum  of  the  measures 
of  zi  1  and  2. is  a  semicircumference,  or 
two  quadrants.] 

361.  Exercise.  The  sides  of  an  inscribed  angle  intercept  an  arc  of 
50°.     What  is  the  size  of  the  angle  ? 

362.  Exercise.  How  many  degrees  in  an  arc  intercepted  by  the 
sides  of  an  inscribed  ajigle  of  40°  ? 

363.  Exercise.  If  the  opposite  angles  of  a  quadrilateral  are  supple- 
mentary, a  circle  may  be  circumscribed  about  it.     (Converse  of  Cor.  V.) 

[Pass  a  circumference  through  three  of  the  vertices.  Then  show  that 
the  fourth  vertex  can  fall  neither  without  nor  within  the  circumference.] 

364.  Exercise.  Show  by  §  355  that  the  sum  of  the  angles  of  a 
triangle  is  two  right  angles. 

365.  Exercise.  Any  parallelogram  inscribed  in  a  circle  is  a 
rectangle. 

366.  Exercise.  Two  circles  are  tangent  at  A. 
AD  and  AE  are  drawn  through  the  extremities  of 
a  diameter  BG. 

Prove  that  DE  is  also  a  diameter.  I    ^/ 

367.  Exercise.  Prove  the  preceding  exercise 
when  the  two  circles  are  tangent  externally.  ^ 

368.  Exercise.  The  angles  of  an  inscribed  trapezoid  are  equal  two 
and  two. 

369.  Exercise.     Prove  §  355,  Case  I,  using  the  figure  of  §  322. 

370.  Exercise.  Two  chords  AB  and  CD  intersect  in  a  circle  at  the 
point  E.  Their  extremities  are  joined  by  the  lines  AC  and  DB.  Prove 
the  A  ACE  and  BDE  mutually  equiangular. 

371.  Exercise.  The  sum  of  one  set  of  alternate  angles  of  an  inscribed 
octagon  is  equal  to  the  sum  of  the  other  set. 

Sanders'  geom.  —  8 


114 


PLANE   GEOMETRY 


Proposition  XXI.     Theorem 

372.  An  angle  formed  hy  two  intersecting  chords  is 
measured  by  one  half  the  sum  of  the  arc  intercepted  hy 
the  sides  of  the  angle  and  the  arc  intercepted  hy  the 
sides  of  its  vertical  angle. 

Let  Zl  be  an  angle  formed  by  the  intersecting  chords 
AB  and  CD. 

To  Prove  Z 1  ~  i  (^D  +  5C). 

Proof.     Draw  the  chord  AC. 

Z1  =  Z2  +  Z3.     (?) 

Z.2^^BC.     (?) 

Z3~i^i).     (?) 

Since  Z 1  is  the  sum  of  ^  2  and  3,  it 
is  measured  by  the  sum  of  their  measures, 

.'.  Z\^\{AD -\- EC).  Q.E.D. 

373.  Exercise.     Derive  the  measure  of  Z  4  in  the  above  figure. 

374.  Exercise.     If  in  the  above  figure  the  arc  C 
BG  contains  124°  and  the  arc  AD  contains  172°, 
how  many  degrees  in  Z 1  ? 

375.  Exercise.      Prove     /.l-^l^AC  +  BD), 
using  this  figure. 

{DE  is  drawn  parallel  to  AB.~\ 


376.  Exercise.  If  angle  1  (figure  §  375)  contains  85°  and  arc  BG 
contains  55°,  how  many  degrees  in  the  arc  AD  ? 

377.  Exercise.  Four  points  A,  B,  G,  and  D  are  so  taken  in  a  cir- 
cumference that  the  arcs  AB,  BG,  GD,  and  DA  form  a  geometrical  pro- 
gression (AB  =  2  BG,  BG  =  2  GD,  etc.).  Find  the  values  of  each  of  the 
angles  formed  by  the  intersection  of  the  chords  AC  and  BD. 


BOOK  II  115 

Proposition  XXII.     Theorem 

378.  An  angle  formed  hy  a  chord  jneeting  a  tangent 
at  the  point  of  tangency  is  measured  hy  one  half  the 
arc  intercepted  hy  its  sides. 

Let  Zl  be  an  angle  formed  by  the  chord  AB  and  the  tan- 
gent CD. 

To  Prove  Z 1  ~  i  amb. 

Proof.     Draw  the  diameter  EB 
to  the  point  of  tangency. 

Z^5(7  =  1R.A.     (?) 

A 
A  right  angle  is  measured  by  a 

quadrant.     (?) 

\  arc  EMB  is  a  quadrant.    (?) 

Zebc^^emb. 

ZEBAr^\AE.     (?) 

Zl,  which  is  the  difference  between  Zebc  and  Zeba,  is 
measured  by  the  difference  of  their  measures. 

Zl^^EMB  —^EA. 

Zl^^AMB.  Q.E.D. 

Similarly,  it  may  be  shown  that  Z  ABD,  which  is  the  sum  of 
R.A.  EBB  and  Z  EBA,  is  measured  by  the  sum  of  their  meas- 
ures, which  is  I"  arc  AEB. 

379.  Exercise.  A  chord  that  divides  a  circumference  into  arcs  con- 
taining 80"^  and  280",  respectively,  is  met  at  one  extremity  by  a  tangent. 
What  are  tlie  angles  formed  by  the  lines  ? 

380.  Exercise.  A  chord  is  met  at  one  extremity  by  a  tangent,  mak- 
ing with  it  an  angle  of  55°.  Into  what  arcs  does  the  chord  divide  the 
circumference  ? 


116 


PLANE    GEOMETRY 


381.  Exercise.  If  two  circles  are  tangent  either  externally  or  in- 
ternally, and  through  the  point  of  contact  two  lines  are  drawn  meeting 
one  circumference  in  B  and  D  and  the  other  in  E  and  C,  BD  and  EC  are 
parallel. 


[Draw  the  common  tangent  mn.  Show  that  Z.3  and  Z2  each  equals 
Zl.] 

382.  Exercise,  If  tangents  be  drawn  to  the  two  circles  at  the  points 
B  and  C  (see  the  figures  of  the  preceding  exercise),  prove  they  are 
parallel. 

Propositiox  XXIII.     Theorem 

383.  An  angle  formed  hy  two  secants  meeting  without 
the  circle  is  measured  by  one  half  the  difference  of  the 
arcs  intercepted  hy  its  sides. 


Let  Z  1  be  an  angle  formed  by  the  two  secants  AB  and  CB. 
To  Prove  Z 1  ~  i  (.ic  -  be). 

Proof.   Draw  the  chord  CE. 

Z1=Z2-Z3.     (?) 
Z  1  is  therefore  measured  by  the  difference  of  the  measures 
of  Z  2  and  3,  i.e.  by  ^{AC  —  BE).  q.e.d. 


BOOK  II 


117 


384.  Exercise.  If  the  secants  AB 
and  CB  in  the  figure  of  §  383  intercept 
arcs  of  70°  and  42°,  what  is  the  size  of    p. 


385.   Exercise.    Prove  §  383, 
this  figure.     [Di^is  II  to  ^C] 


using 


Proposition  XXIY.     Theorem 

386.  An  angle  formed  by  a  tangent  and  a  secant  meet- 
ing without  the  circle  is  measured  by  one  half  the  dif- 
ference of  the  arcs  intercepted  by  its  sides. 


Let  Zl  be  an  angle  formed  by  the  tangent  AB   and  the 
secant  CB. 
To  Prove  Z  1  ~  i  (J  C  —  AD). 

Proof.    Similar  to  that  of  §  383. 


Exercise.  Prove  §  386,  using  this 
figure.     \_EA  is  II  to  ^C] 

387.  Exercise.  A  tangent  and  a 
secant  meeting  without  a  circle  form  an 
angle  of  35°.  One  of  the  arcs  intercepted 
by  them  is  15°.  How  many  degrees  in 
the  other  ? 

388.  A  triangle  ABC  is  inscribed  in  a  circle.  The  angle  B  is  equal  to 
50°,  and  the  angle  C  is  equal  to  60°.  What  angle  does  a  tangent  at  A 
make  with  BC  produced  to  meet  it  ? 


118 


PLANE   GEOMETRY 


Proposition  XXV.     Theorem 

389.  An  angle  formed  hy  two  tangents  is  measured 
hy  one  half  the  difference  of  the  arcs  intercepted  hy  its 
sides. 


Let  Z  1  be  an  angle  formed  by  the  tangents  AB  and  CB. 
To  Prove  Z  1  ~  i  {ANC  —  AMC). 

Proof.     Similar  to  that  of  §§  383  and  386. 


Exercise.     Prove  §  389,  using  this       D^ 
figure. 

IAD  is  drawn  parallel  to  ^0.  ]  N 


Exercise.     Prove  §  389,  using  this 


figure. 


[BD  is  any  secant  drawn  from  2?.] 


390.    Exercise.     The  angle  formed  by  two  tangents  is  74°.      How 
many  degrees  in  each  of  the  two  arcs  intercepted  by  them  ? 


BOOK   II 


119 


Proposition  XXVI.     Problem 

391.    Through  a  given  point  to  draw  a  tangent  to  a 
Hven  circle. 

Case  I 


When  the  given  point  is  on  the  circumference. 

Let  A  be  the  given  point  on  the  circumference  of  the  circle 
whose  center  is  0. 

Required  to  draw  a  tangent  to  the  circle  through  A. 

See  §  307. 

Case  II 

When  the  given  point  is  with- 
out the  circumference. 

Let  A  be  the  given  point  with- 
out the  circle  whose  center  is  0. 

Required  to  draw  a  tangent  to 
the  circle  through  A. 

Draw  OA. 

On  OA  as  a  diameter,  describe 
a  circumference,  cutting  the  given  circumference  at  B  and  C. 

Draw  AB  and  AC. 

AB  and  AG  are  the  required  tangents. 

Draw  the  radii  OB  and  OC. 

Z  1  is  a  right  angle.     (?) 

AB  is  tangent  to  the  circle.     (?) 

Similarly,  ^C  is  tangent  to  the  circle,  q.e.f 


120 


PLANE    GEOMETRY 


B 


E 


Case  II.     Second  Method 

392.  With  A  as  center  and  ^0  as  a  radius,  describe  the 
arc  DE. 

With  O  as  a  center  and 
the  diameter  of  the  given 
circle  as  a  radius,  describe 
an  arc  cutting  DE  2A,  B. 

Draw  OB  intersecting 
the  given  circle  at  C. 

Draw  AC.  Then  ^C  is 
the  required  tangent. 

[The  proof  is  left  for 
the  student.] 

393.  Corollary.  The  two  tangents  drawn  from  a  point  to  a 
circle  are  equal ;  and  the  line  joining  the  point  with  the  center  of 
the  circle  bisects  the  angle  between  the  tangents,  and  also  bisects  the 
chord  of  contact  (B  C  in  the  figure  to  first  method)  at  right  angles. 

394.  Scholium.  When  the  given  point  is  without  the  circle, 
two  tangents  can  be  drawn ;  when  it  is  on  the  circumference, 
one,  and  when  it  is  within  the  circle,  none. 

395.  Definition.  A  polygon  is  circumscribed  about  a  circle 
when  each  of  its  sides  is  tangent  to  the  circle.  In  this  case 
the  circle  is  said  to  be  inscribed  in  the  polygon. 

396.  Exercise.  If  a  quadrilateral  is  circumscribed  about  a  circle,  the 
sum  of  one  pair  of  opposite  sides  is  equal  to  the  sum  of  the  other  pair. 

Suggestion.     Use  §  393. 

397.  Exercise.  From  the  point  A 
two  tangents  AB  and  AC  are  drawn  to 
the  circle  whose  center  is  0. 

At  any  point  D  on  the  included  arc  BC^ 
a  third  tangent  FE  is  drawn. 

Prove  that  the  perimeter  of  the  A  AEF 
is  constant,  and  equal  to  the  sum  of  the 
tangents  AB  and  AC, 


BOOK  II 


121 


398.  Exercise.  To  inscribe  a 
circle  in  a  given  triangle. 

Bisect  two  of  the  angles.  Show 
that  their  point  of  meeting  is  equally 
distant  from  the  three  sides. 

.-.  the  three  perpendiculars  01, 
02,  and  03  are  equal. 

With  0  as  a  center  and  with  0  1 
as  a  radius,  describe  the  required 
circle. 

Proposition  XXYII.     Problem 

399.  On  a  given  line  to  construct  a  segment  that  shall 
contain  a  given  angle. 


Let  AB  hQ  the  given  line  and  Z  M  the  given  angle. 

Required  to  construct  on  AB  a  segment  that  shall  contain  Z  M. 

Draw  CD  through  J5,  making  /.I  =  /.  M.' 

Erect  BE  ±  to  CD  and  bisect  AB  hy  the  J_  FG. 

Prove  that  BE  and  FG  meet  at  some  point  0. 

Show  that  0  is  equally  distant  from  A  and  B. 

With  0  as  a  center  describe  a  circle  passing  through  A  and  B. 

X>C  is  tangent  to  this  circle.     (?)         Z  1  ~  ^  AB.     (?) 

Inscribe  any  angle  as  /.  ASB  in  the  segment  ARB. 

ZASB^^AB.      (?)  ZASB  =  Zl  =  Z3f.      (?) 

The  segment  ARB  is  the  required  segment,  since  any  angle 
inscribed  in  it  is  equal  to  Z  M.  q.e.f. 


122 


PLANE    GEOMETRY 


400.  Exercise.     On  a  given  line  construct  a  segment  that  shall  con- 
tain an  angle  of  135°. 

401.  Exercise.     What  is  the  locus  of  the  vertices  of  the  vertical 
angles  of  the  triangles  having  a  common  base  and  equal  vertical  angles  ? 

402.  Exercise.     Construct   a  triangle,  having  given   the   base,  the 
vertical  angle,  and  the  altitude. 

403.  Exercise.     Construct  a  triangle,   having  given  the  base,  the 
vertical  angle,  and  the  medial  line  to  the  base. 


EXERCISES 


1.  Two  secants,  AB  and  AC^ 
are  drawn  to  the  circle,  and  AB 
passes  through  the  center.  A 

Prove       AB>AC. 


2.  One  angle  of  an  inscribed  triangle  is  42°,  and  one  of   its  sides 
subtends  an  arc  of  110°. 

Find  the  angles  of  the  triangle. 

3.  Two  chords  drawn  perpendicular  to  a  third 
chord  at  its  extremities  are  equal.  [Show  tliat 
BC  and  AD  are  diameters,  and  that  k^ABC  and 
ABB  are  equal] 


4.    AB  and  CD  are  two  chords  intersecting  at 
E,  and  CE=BE. 

Prove  AB  =  CD. 


5.    ABC  is  a  triangle  inscribed  in  the  circle, 
whose  center  is  O. 

OD  is  drawn  perpendicular  to  AC. 
Prove  ADOC^-AB. 


BOOK  II 


123 


6.    What  is  the  locus  of  the  centers  of  circles  tangent  to  a  line  at  a  given 
point  ? 


7.  P  is  any  point  within  the  circle  whose  center 
is  0.  Prove  that  PA  is  the  shortest  line  and  PB 
the  longest  line  from  P  to  the  circumference. 


8.  If  a  circle  is  described  on  the  radius  of  an- 
other circle  as  a  diameter,  any  chord  of  the  greater 
circle  drawn  from  the  point  of  contact  is  bisected 
by  the  circumference  of  the  smaller  circle. 


9.    If  from  a  point  on  a  circumference  a  number  of  chords  are  drawn, 
find  the  locus  of  their  middle  points.     (Ex.  8.) 

10.  From  two  points  on  opposite  sides  of  a  given  line,  draw  two  lines 
meeting  in  the  given  line,  and  making  a  given  angle  with  each  other. 
(§399.) 

11.  Work  Ex.  10,  taking  the  two  points  on  the  same  side  of  the  given 
line. 

When  is  the  problem  impossible  ? 

12.  One  of  the  equal  sides  of  an  isosceles  triangle 
is  the  diameter  of  a  circle. 

Prove  that  the  circumference  bisects  the  base. 
[Show  that  BD  is  ±  to  AC] 

13.  What  is  the  locus  of   the  centers  of  circles 
having  a  given  radius  and  tangent  to  a  given  line  ? 

14.  Describe  a  circle  having  a  given  radius  and  tangent  to  two  non- 
parallel  lines. 

How  many  circles  can  be  drawn  ? 

15.  What  is  the  locus  of  the  centers  of  circles  having  a  given  radius 
and  tangent  to  a  given  circle  ? 

16.  Describe  a  circle  having  a  given  radius  and  tangent  to  two  given 
circles. 


124 


PLANE   GEOMETRY 


17.    Describe  a  circle  having  a  given  radius  and  tangent  to  a  given  line 
and  also  to  a  given  circle. 

C 


18.  The  base  AB  of  the  isosceles  triangle  ABG 
is  a  chord  of  a  circle,  the  circumference  of  which 
intersects  the  two  equal  sides  at  D  and  E. 

Prove  CD  =  CE.  A 

[Z  A  and  Z  B  are  measured  by  equal  arcs.  ] 


19.  If  an  isosceles  triangle  is  inscribed  in  a  cir- 
cle, prove  that  the  bisector  of  the  vertical  angle 
passes  through  the  center  of  the  circle.  Al 


20.  The  altitude  of  an  equilateral  triangle  is  one 
and  a  half  times  the  radius  of  the  circumscribed 
circle. 

[Use  the  preceding  Exercise  and  §  245.] 


21.  If  a  triangle  is  circumscribed  about  a 
circle,  the  bisectors  of  its  angles  pass  through 
the  center  of  the  circle.     [§  230.] 


22.  The  altitude  of  an  equilateral  triangle  is  three  times  the  radius  of 
the  inscribed  circle. 

[Use  Ex.  21  and  §245.] 

23.  The  angle  between  two  tangents  to  a  circle  is  30°.      Find  the 
number  of  degrees  in  each  of  the  intercepted  arcs. 


BOOK  II 


125 


24.  From  a  given  point  draw  a  line 
cutting  a  circle  and  making  the  chord 
equal  to  a  given  line. 

[The  chord  B8  \q  equal  to  the  given 
line.  The  dotted  circle  is  tangent  to 
R8.-\ 


25.  Find  the  angle  formed  by  two 
tangents  to  a  circle,  drawn  from  a  point 
the  distance  of  which  from  the  center  of 
the  circle  is  equal  to  the  diameter. 


26.  With  a  given  radius  describe  a  circle  that  shall  pass  through  a 
given  point  and  be  tangent  to  a  given  line. 

27.  With  a  given  radius  describe  a  circle  that  shall  pass  through  a 
given  point  and  be  tangent  to  a  given  circle. 


28.    From  a  point  without  a  circle   draw  the  shortest  line  to  the 
circumference. 


29.  ABC  is  an  inscribed  equilateral  triangle. 
DE  joins  the  middle  points  of  the  arcs  BC  and 
CA.  Prove  that  DE  is  trisected  by  the  sides  of 
the  triangle.  /V 


30.  Find  a  point  within  a  triangle  such  that 
the  angles  formed  by  drawing  lines  from  it  to  the 
three  vertices  of  the  triangle  shall  be  equal  to  each 
other.     (§  399.) 


126 


PLANE    GEOMETRY 


31.  A  median  BD  is  drawn  from  angle  B  in 
the  triangle  ABC.  Show  that  angle  5  is  a  right 
angle  when  BD  is  equal  to  one  half  of  the  base 
ACy  an  acute  angle  when  BD  is  greater  than  one 
half  of  AC,  and  an  obtuse  angle  when  BD  is  less 
than  one  half  of  AC. 


32.  In  any  right-angled  triangle,  the  sum  of 
the  two  legs  is  equal  to  the  sum  of  the  hypote- 
nuse and  the  diameter  of  the  inscribed  circle. 

[Tangents  drawn  from  a  point  to  a  O  are 
equal.] 


33.  Tangents  CA  and  DB  drawn  at  the  ex- 
tremities of  the  diameter  AB  meet  a  third  tangent 
CD  at  C  and  D.     Draw  CO  and  DO. 

Prove   CD=CA  +  DB    and  Z  COD  =  1  R.A. 


34.  If  from  one  point  of  intersection 
of  two  circles  two  diameters  are  drawn, 
the  other  extremities  of  the  diameters 
and  the  other  point  of  intersection  of  the 
circles  are  in  a  straight  line. 

[Draw  DE  and  EF.  Show  that 
/.DEC -Y  ACEF=2n.A.'B.^ 

35.  Through  the  points  of  intersec- 
tion of  two  circles  two  parallel  secants 
are  drawn,  terminating  in  the  curves. 
Prove  the  secants  equal. 

[Show  that  the  quadrilateral  ECDF 
has  its  opposite  angles  equal,  each  to 
each.  ] 

36.  In  a  given  circle  draw  a  chord  the  length  of  which  shall  be  twice 
its  distance  from  the  center. 


BOOK  II 


127 


37.  Three  equal  circles  are  tangent  to  each  other.  Through  their 
points  of  contact  three  oommon  tangents  are  drawn. 

Prove.     1.    The  three  tangents  meet  in  a  common  point. 

2.    The  point  of  meeting  is  equally  distant  from  the  three 
points  of  contact. 

38.  The  sum  of  the  angles  subtended  at  the  center 
of  a  circle  by  two  opposite  sides  of  a  circumscribed 
quadrilateral  is  equal  to  two  right  angles. 

[To  prove  /.AOB-^  Z.  COD  =  2  R.A.'s.] 

39.  Find  the  locus  of  points  such  that  tangents 
drawn  from  them  to  a  given  circle  shall  equal  a  given  . , 
line. 

40.  Inscribe  a  circle  in  a  given  quadrant. 
[OD  bisects  ZAOB.     DE  is  ±  to  OB.     DF 

bisects  Z.  ODE.'] 

41.  If  the  tangents  to  a  circle  at  the  four  verti- 
ces of  an  inscribed  rectangle  (not  a  square)  be 
prolonged,  they  form  a  rhombus. 

42.  From  any  point  (not  the  center)  within  a  circle  only  two  equal 
straight  lines  can  be  drawn  to  the  circumference. 

43.  Given  a  circle  and  a  point  within  or  without  (not  the  center), 
using  the  given  point  as  a  center  to  describe  a  circle,  the  circumference 
of  which  shall  bisect  the  circumference  of  the  given  circle. 

44.  In  a  given  circle  inscribe  a  triangle,  the  angles  of  which  are  respec- 
tively equal  to  the  angles  of  a  given  triangle. 

[Draw  a  tangent  to  the  O,  and  from  the  point  of  contact  draw  two 
chords,  making  the  three  A  at  the  point  of  contact  equal  to  the  A  of 
the  A.] 

45.  Circumscribe  about  a  given  circle  a  triangle,  the  angles  of  which 
are  respectively  equal  to  the  angles  of  a  given  triangle. 

[Inscribe  a  O  in  the  given  A.] 

46.  Of  all  triangles  having  a  common  base 
and  an  equal  altitude,  the  isosceles  triangle  has 
the  greatest  vertical  angle. 

47.  Given  the  base,  the  vertical  angle,  and 
the  foot  of  the  altitude,  construct  the  triangle. 


128 


PLANE   GEOMETRY 


48.  If  the  sum  of  one  pair  of  opposite  sides 
of  a  quadrilateral  is  equal  to  the  sum  of  the 
other  pair,  a  circle  can  be  inscribed  in  the 
quadrilateral. 

[Describe  a  O  tangent  to  three  of  the  sides. 
Show,  by  §  396,  that  the  fourth  side  can  neither 
cut  this  circle  nor  lie  without  it.] 

49.  Any  point  on  the  circumference  circum- 
scribing an  equilateral  triangle  is  joined  with 
the  three  vertices. 

Prove  that  the  greatest  of  the  three  lines  is 
equal  to  the  sum  of  the  other  two. 

[Lay  off  DE  =  DC.  Prove  ^  AEC  and 
^Z>C equal  in  all  respects.] 

50.  Two  equal  circles  intersect  at 
A  and  B.  On  the  common  chord  AB 
as  a  diameter  a  third  circle  is  described. 
Through  A  any  line  CD  is  drawn  ter- 
minating in  the  circumferences  and  in- 
tersecting the  third  circumference  at  E. 

Prove  that  CD  is  bisected  at  E. 
[Show  that  A  BCD  is  isosceles,  and 
that  BE  is  ±  to  the  base  (7Z>.] 

51.  Two  equal  circles  intersect  at 
A  and  B.  With  ^  as  a  center,  any 
circle  is  described  cutting  the  two 
equal  circumferences  at  G  and  D. 

Prove  that  J.,  C,  and  D  are  in  a 
straight  line. 

[Drawee.  ZBAC'^^BC.  But 
BC=BD.  Draw  ^2).  ZBAD^^BD. 
.-.  ZBAC  =  ZBAD.^ 

62.  If  two  circles  intersect,  the 
longest  common  secant  that  can  be 
drawn  through  either  point  of  inter- 
section is  parallel  to  the  line  joining 
their  centers. 

rShow  that  CD  =  2AB,  and  that 
any  otner  secani,  through  E  is  less  than 
2AB.2 


BOOK  III 


404.  Definitions.     A  proportion  is  the  equality  of  ratios. 

-  =  -  is  a  proportion,  and  expresses  the  fact  that  the  ratio  of 
b     d 

a  to  6  is  equal  to  the  ratio  of  c  to  d.  The  proportion  -  =  - 
may  also  be  written  a:b=c:d  and  a:b  ::  c:d. 

In  the  proportion      =  -,  the  first  and  fourth  terms  (a  and  d) 

0         Oj 

are  called  the  extremes^  and  the  second  and  third  terms  (b  and 
c)  are  called  the  means.  The  first  and  third  terms  (a  and  c) 
are  the  antecedents,  and  the  second  and  fourth  terms  (6  and  d) 
are  the  consequents. 

In  the  proportion   -  =  -,  d  is  called  a  fourth  proportional  to 
b     d 

the  three  quantities  a,  b,  and  c. 

If  the  means  of  a  proportion  are  equal,  either  mean  is  a 
mean  proportional  or  a  geometrical  mean  between  the  extremes. 

Thus  in  the  proportion  -  =  -,  6  is  a  mean  proportional  between 
b      c 

a  and  c.  In  this  same  proportion,  c  is  called  a  third  pi'opor- 
tional  to  a  and  b. 

Proposition  I.     Theorem 

405.  In  a  proportion,  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

M-  (^> 

To  Prove  ad  =  be. 

Proof.     [Clear  fractions  in  (1)  by  multiplying  both  members 
by  6d]  Q.E.D. 

SANDERS'  GEOM.  — 9  129 


130  PLANE   GEOMETRY 

406.  Corollary.     The  mean  proportional  between  two  quan- 
tities is  equal  to  the  square  root  of  their  product. 

Let  ^  =  ^.  (1) 

be  ^ 

To  Prove  b=zVac. 

Proof.     [Apply  §  405  to  (1),  and  extract  the  square  root  of 
both  members.]  q.e.d. 

407.  Exercise.     Find  x  in  —  =  —. 

12      X 

408.  Exercise.     What  is  the  geometrical  mean  or  mean  proportional 
between  9  and  4  ? 

409.  Exercise.     12  is  the  geometrical  mean  between  two  numbers. 
One  of  them  is  16.     What  is  the  other  ? 

410.  Exercise.     Find  the  mean  proportional  between  a^  -{■  2  ab  +  b^ 
and  a^-2ab  +  b^. 


Proposition  II.     Theorem.     (Converse  of  Prop.  I.) 

411.  If  the  product  of  two  factors  is  equal  to  the  prod- 
uct of  two  other  factors,  the  factors  of  either  product 
may  he  made  the  means,  and  the  factors  of  the  other 
product  the  eoctremes  of  a  proportion. 

Let  ad  =  bc.  (1) 

To  Prove  -=-. 

6      d 

Proof.     [Divide  both  members  of  (1)  by  bd."]  q.e.d. 

412.  Exercise.  From  the  equation  ad  =  be,  derive  the  following 
eight  proportions. 

a_c  ^  —  b  ^  =  ^  ^  =  ^ 

b     cC  c     d!  d     b^  a     b*       ' 

b_d  b_a  ^  —  b  '^  —  ^ 

a     c  d     c  c     a  b     a 


BOOK  III  131 

413.  Exercise.     Form  different  proportions  from 

414.  Exercise.     Form  a  proportion  from 

a2  +  2  a&  +  62  ^  r^^^y^ 

Wiiat  is  a  +  6  called  in  this  proportion  ? 

415.  Exercise.     Form  a  proportion  from  a^  +  6^  =  x^  —  y^. 

416.  Definition.  A  proportion  is  arranged  by  alternation 
when  antecedent  is  compared  with  antecedent  and  consequent 
with  consequent. 

If  the  proportion  -  =  -  is  arranged  by  alternation,  it  be- 

a     b 
conies  -  =  — 
c      d 

Proposition  III.     Theorem. 

417.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  alternation. 

Let  -  =  -.  (1) 

To  Prove  -  =  -. 

c      d 

Proof.     Apply  §  405  to  (1)  ad  =  he.  (2) 

Apply  §  411  to  (2)     ^=^' 

418.  Exercise.  Write  a  proportion  that  will  not  be  altered  when 
arranged  by  alternation. 

419.  Definition.  A  proportion  is  arranged  by  inversion 
when  the  antecedents  are  made  consequents,  and  the  conse- 
quents are  made  antecedents. 

If  the  proportion   -  =  -    is  arranged  by  inversion,  it  be- 

b  d         ^  '^ 

comes  —  =  — 
a     c 


132  PLANE    GEOMETRY 


Proposition  IV.     Theorem 

420.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  inversion. 

To  Prove  -=-. 

a      c 

Proof.     Apply  §  405  to  (1)  ad  =  be.  (2) 

Apply  §  411  to  (2)     l  =  f 

421.  Definition.  A  proportion  is  arranged  by  composition 
when  the  sum  of  antecedent  and  consequent  is  compared  with 
either  antecedent  or  consequent. 

The  proportion  -  =  -  arranged  by  composition  becomes 

a  +  b  _c-\-  d         a  +  b  _c  +  d 
a  c  b  d 


Proposition  V.     Theorem 

422.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  hy  composition. 

1=1-  (^> 

To  Prove  a  +  b^c  +  d^ 

a  c 

Proof.     Apply  §  405  to  (1) 

ad  =  be.  (2) 

Add  ac  to  both  members  of  (2) 

oc  +  ad  =  ac  -f  be.  (3) 

Factor  (3)  a(c -\- d)  =  e(a -\-  b).  (4) 

Apply  §  411  to  (4) 

a-\-b  _c  -\-  d 

a       ""       C      *  Q.E.D. 


BOOK  III  •  133 

423.  Note.  The  student  may  discover  for  himself  the  steps  of  the 
solution  of  this  and  the  succeeding  propositions  by  studying  the  analysis 
of  the  theorem. 

In  the  analysis  we  assume  the  conclusion  (the  part  to  be  proved)  to  be 
a  true  equation.  Working  upon  this  conclusion  by  algebraic  transforma- 
tions, we  produce  the  hypothesis. 

The  solution  of  the  theorem  begins  with  the  last  step  of  the  analysis 
dJi^L  reverses  the  work,  step  by  step,  until  the  first  step  or  conclusion  is 
reached. 

In  §  422  we  have  given  ^  =  -•  (1) 

h     d 

We  are  to  prove  ^-±-^  =  ^^ti.  (2) 

a  c 

Analysis 
Clear  fractions  in  (2)     c{a -\-  b)  =  a{c  +  d).  (3) 

Expand  (3)  ac  -{■  be  =  ac  -\-  ad.  (4) 

Subtract  ac  from  both  members  of  (4). 

be  =  ad.  (5) 

Apply  §  411  to  (5)  l  =  t  (6) 

b     d 

Let  the  student  show  that  the  solution  of  Prop.  V.  as  given  on  the  pre- 
ceding page  may  be  obtained  by  reversing  the  steps  of  this  analysis. 

424.  Exercise.     Let  -=-• 

b     d 

To  Prove  a±b^c±d^ 

b  d 

425.  Exercise.     Arrange  ^  ~  ■  =  ^-^^ —  by  composition. 

b  d 

426.  Exercise.     Arrange  ^—  =  — — —  by  composition  and  then 

4  X 

find  the  value  of  x. 

427.  Definition.  A  proportion  is  arranged  by  division 
when  the  difference  between  antecedent  and  consequent  is  com- 
pared with  either  antecedent  or  consequent. 

The  proportion  -  =  -  arranged  bv  division  becomes 
b      d 

a  —  b      c  —  d      a  —  b      c  —  d      b  —  a      d  —  c      b  —  a      d  —  c 
= or = or = or —  = • 


134  PLANE   GEOMETRY 


Proposition  VI.     Theorem 

428.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  by  division. 

Let  -  =  -.  m 

h     d  ^^ 

To  Prove  ^!^~^  =  ^-^-  (2) 

a  c  ^ 

Proof.     [_Analysis.     Clear  fractions  in  (2) 

c(a  —  b)=a(c  —  d),  (3) 

Expand  (3)  ac  —  bc  =  ac  —  ad.  (4) 

Subtract  ac  from  both  members  of  (4) 

—  bc  =  —  ad.  (5) 

Divide  both  members  of  (5)  by  —  1 

be  =  ad.  (6) 

Apply  §  411  to  (6)  1  =  ^1.  Q.E.D. 

Let  the  student  derive  the  solution  of  Prop.  VI.  from  the 
analysis. 

42a.    Exercise.     If     a  +  b-c^a^:^ 
c-^-d  +  a       2d 

then  *  _±_^a  +  c-d^ 

a  —  c  2d 

430.  Definition.  A  proportion  is  arranged  by  composition 
and  division,  when  the  sum  of  antecedent  and  consequent  is 
compared  with  the  difference  of  antecedent  and  consequent. 

The  proportion  -  =  -,  arranged  by  composition  and  division, 
b     d 

becomes  a^f)^c-\-d 

a  —  b     c  —  d 


BOOK  III  135 


Proposition  VII.     Theorem 

431.  If  four  quantities  are  in  proportion,  they  are  in 
proportion  hy  composition  and  division. 

Let  "      ' 


To  Prove 


h     d 

a-\-h      c  +  d 


a  —  h      c  ~  d 
Proof.     [Analyze  and  solve.] 

432.    Exercise.     If  ^  =  ^, 

b     d 

««^„^  c  —  a     d  —  b 
prove 


a  +  c     b  +  d 


Proposition  VIII.     Theorem 

433.  If  four  quantities  are  in  proportion,  like  powers 
of  those  quantities  are  proportional. 

M  <^) 

To  Prove  ^  =  £l. 

Proof.     [Raise  both  members  of  (1)  to  the  nth  power.]    q.e.d. 

434.  Corollary.     If  four  quantities  are  in  proportion^  like 
roots  of  those  quantities  are  propoi'tional. 


.    Exercise. 

If 

a  _ 
b~ 

'd' 

show  that  ^  = 

a2 

02. 

-62 

-d^ 

I.    Exercise. 

If 

a 
b 

c 

'd' 

show  that 

a3  +  63  _ 

.  c3  +  # 

qS  —  ^3      c^  —  d* 


136  PLANE    GEOMETRY 

Proposition  IX.     Theorem 

437.  If  four  quantities  are  in  proportion,  equimul- 
tiples of  the  antecedents  are  proportional  to  equimul- 
tiples of  the  consequents. 

Let  ^  =  ^.  (1) 

To  Prove 

hy     dy 

Proof.     Multiply  both  members  of  (1)  by  -•  q.,e.d. 

438.  Exercise.     Let 
To  Prove 

439.  Exercise.    Let 


To  Prove  ab  +  cd^a^  +  c\ 

ah  —  cd     a^  —  c^ 


a 
h' 

_  c 
~d' 

ax 
by" 

ex 
'dy 

embers  c 

a 
h~ 

c 

'  d 

ac 
bd 

C2 

a  _ 
b~ 

'  d 

440.    Exercise. 

Let 

a_b 
b      c" 

To  Prove 

a  +  c  _  62  -j_  c2 
a  -  c  ~  62  _  c2 

441.    Exercise. 

Let 

a      c 
b~d 

To  Prove 

ma^  +  wc2  _  a2 
W&2  +  nd2      62 

442.  Definition.  A  continued  proportion  is  a  proportion 
made  up  of  several  ratios  that  are  successively  equal  to  each 
other.     Example : 

«  =  ^=  «  =2,  etc. 
b     d     f     h' 


BOOK  III 


137 


Propositiox  X.     Theorem 

443.  In  a  continued  proportion  the  sum  of  the  ante- 
cedents is  to  the  sum  of  the  consequents  as  any  antecedent 
is  to  its  consequent. 

(1) 


Let 

a  _ 

_c  _ 

f     h 

To  Prove 

a-[-  c  +  e  -\-  g  _e 
h+d+f+h     f 

Proof 

a      e 

(2) 

c      e 
e       e 

7"7 

(3) 
(4) 

'  From  ( 

h      f 

(5) 

af=be 

(6) 

cf=de 
ef=fe 

(8) 

From 

gf=he 

(9). 

From  (2),  (3),  (4),  and  (5). 


Add  (6),  (7),  (8),  and  (9),  and  factor. 

/(a  +  c  +  e  +  g)=eiib  +  d  +/+  h). 

Apply  §  411  to  (10). 

a-^c  +  e  +  g^e 
b  +  d-\-f+h     f 


(10) 


Q.E.D. 


444.    Exercise,     If 


then  will 


x  +  y 

a  +  h 


_V_z 

y  +  z  _z  +  x^ 
b  -\-  c     c  +  a 


445.    Exercise.     Let 


To  Prove 


a  _  c  _  ^  _  flf 

b~d~  f~  h 

a  —  c-\-  e  —  g _  c 

b-d-\-f-h~  d 


138  PLANE    GEOMETRY 

Proposition   XI.     Theorem 

446.  If  the  terms  of  one  proportion  are  multiplied  hy 
the  corresponding  terms  of  another  proportion,  the  prod- 
ucts are  proportional. 

Let  ^=£     (1)     and      ?  =  ^     (2). 

m    Tk-.  ax      cm 

To  Prove        —  =  — 

hy      dn 

Proof.     [The  proof  is  left  to  the  student.]  q.e.d. 

447.  Exercise.  If  the  terms  of  one  proportion  are  divided  by  the 
corresponding  terms  of  another  proportion,  the  quotients  are  proportional. 

448.  Exercise.     If  -  = -, 

h     d 

show  that  ^'  +  «^^  +  ^'  =  cl+^d+i^. 
a^  -  ab  -\-  b'^      c-  -  cd  +  d^ 

Proposition   XII.     Theorem 

449.  //  a  number  of  parallels  intercept  equal  distances 
on  one  of  two  transversals,  they  will  intercept  equal  dis- 
tances on  the  otlver  also. 


V  r 

Let  AB^  CD,  EF,  and  G//  be  a  number  of  parallels  cut  by  the 
transversals  xy  and  zr,  making 

AC  =  CE  =  EG. 
To  Prove  BD  =  DF  =  FH. 

Proof.     [Proof  similar  to  that  of  §  240.]  q.e.d. 


BOOK  III  139 

450.  Corollary  I.  A  line  drawn 
from  the  middle  point  of  one  of  the 
inclined  sides  of  a  trapezoid  parallel  to 
either  base,  bisects  the  other  inclined  side. 

451.  Corollary  II.  A  line  join- 
ing the  middle  points  of  the  inclined 
sides  of  a  trapezoid  is  parallel  to  the 
bases. 

Suggestion.      Draw    FG  II  AB.       Prove 
A  CFn  =  A  GDn  whence   Fn  =  nG.     Prove   FG  =  AB 
Prove  AmnG  a  parallelogram. 

452.  Exercise.  A  line  joining  the  middle  points  of  two  opposite 
sides  of  a  parallelogram,  is  parallel  to  the  two  remaining  sides  and  passes 
through  the  point  of  intersection  of  the  diagonals. 

453.  Exercise.  A  line  joining  the  middle  points  of  the  inclined  sides 
of  a  trapezoid  is  equal  to  one  half  the  sum  of  the  parallel  sides. 

[In  the  figure  of  §  451  show  mn  =  ^  {BF 
+  AG)  Sind  CF=GD]. 

454.  Exercise.  If  from  the  extremities  of 
a  diameter  perpendiculars  are  drawn  to  a  line 
cutting  the  circle,  the  parts  intercepted  between 
the  feet  of  the  perpendiculars  and  the  curve  are 
equal. 

[To  prove  CE  =  FD-I 

455.  Exercise.  If  perpendiculars  are  drawn  from  the  extremities  of 
a  diameter  of  a  circle  to  a  line  lying  without  the  circle,  the  feet  of  these 
perpendiculars  are  equally  distant  from  the  center  of  the  circle. 

456.  Exercise.  A  line  joining  the  middle  points  of  the  inclined  sides 
of  a  trapezoid  bisects  the  diagonals  of  the  trapezoid,  and  also  bisects  any 
line  whose  extremities  are  in  the  parallel  bases. 

457.  Exercise.  The  inclined  sides  of  a  trapezoid  are  Oft.  and  15  ft. 
respectively.  If  on  the  shorter  of  these  sides  a  point  is  taken  .S  ft.  from 
one  end,  and  through  that  point  a  parallel  to  either  base  is  drawn,  where 
does  the  parallel  intersect  the  other  inclined  side  ? 


140  PLANE    GEOMETRY 

Proposition  XIII.     Theorem. 

458.   A  line  drawn  parallel  to  one  side  of  a  triangle 
divides  the  other  two  sides  proportionally. 


iet  Z)^  be  parallel  to  BC. 

r^     ^  AD       AE 

To  Prove  —  = 

DB      EC 

Proof.  Case  I.  When  the  segments  AD  and  DB  are  com- 
mensurable. 

Let  the  common  unit  of  measure  be  contained  in  AD  m 
times,  and  in  DB  n  times. 

Whence  i^  =  !^.  (1) 

DB       n 

Divide  AD  into  m  equal  parts,  each  equal  to  the  unit  of 
measure,  and  DB  into  n  equal  parts,  and  through  the  points  of 
division  draw  parallels  to  BC. 

These  parallels  intercept  equal  distances  on  AC  (?).  Conse- 
quently AE  \fi  divided  into  m  equal  parts,  and  EC  into  n  equal 
parts. 

AE     m  ^2) 


q.e.d. 


Whence 

EG      n 

Compare  (1)  and  (2). 

AD       AE 

^ 

DB       EC 

BOOK   III  141 

Case  II.     When  the  segments  AD  and  DB  are  incommen- 
surable. 

A 


Let  DEhQ  parallel  to  BC. 

To  Prove  ^_^  =  4£. 

DB       EC 

Proof.  Divide  AD  into  a  number  of  equal  parts,  and  let  one 
of  these  parts  be  applied  to  Z) 5  as  a  unit  of  measure. 

Since  AD  and  DB  are  incommensurable,  this  unit  of  measure 
will  not  be  exactly  contained  in  DB,  but  there  will  remain  over 
some  distance  MB  smaller  than  the  unit  of  measure. 

Draw  MN  parallel  to  BC. 

Since  AD  and  DM  are  commensurable  (why  ?), 

AD      AE  1      ^        T 
—  =  —  by  Case  I. 

DM      EN 

This  proportion  is  true,  no  matter  how  many  equal  divisions 
are  made  in  AD. 

If  the  number  of  divisions  is  increased,  the  size  of  each 
division  is  diminished,  and  MB  is  also  diminished. 

AD 
As  the  number  of  divisions  is  increased,  the  ratio  is 

approaching  —  as  its  limit,  and  the  ratio  - —  is  approaching 

AE  ^^  ^^ 

—  as  its  limit. 

Since  the  variables  —  and  —  are  always  equal,  and  are 
DM  EN  J        1      J 

each  approaching  a  limit,  their  limits  are  equal  (?). 

mi        £  AD       AE 

Therefore  ^  — = q.e.d. 

DB      EC 


142 


PLANE   GEOMETRY 


459.    Corollary  I.     DE  is  parallel 
to  BC. 

»,     T»_  AD       AE         T    DB       EC 

To  Prove  —  =  —  and  —  =  — ^• 

AB       AC 


Suggestion.    Apply  §  422  to 


AB      AC 

AD^AE 
DB     EG 


460.    Corollary  II.     If  two  lines  are  cut  by  any  number  of 
parallels,  they  are  divided  proportionally. 


Case  I.     When   the   two   lines   are 
parallel. 

MR       RW        WY 


To  Prove 


NS       SX 


xz 


Case    II.     When   the  two   lines   are 
oblique  to  each  other. 


-     T»  AM       MR       RW       WY 

To  Prove    —  =  —  = = 


AN      NS 
Use  §  458  and  §  459. 


SX       XZ 


461.    Corollary  III.     To  construct  a  fourth  proportional  to 
three  given  lines. 

Let  a,  b,  and  c  be  the  three  given  a 

lines.  ^ 

c 

Required  to  construct  a  fourth  pro- 
portional to  them. 

Construct    any    convenient    angle, 
XYZ. 

Lay  off  YD  =  a,  DE  =  b,  and  YF=:c. 

DiSiW  DF.     Draw  EG  II  to  DF. 


BOOK   III 


143 


FG  is  the  required  fourth  proportional. 

c 


YD        YF  ,„.  a 

—  =  —  (?)  or  -  =  — . 

DE       FG      "^         h       FG 


Q.E.F. 
Note,     If  h  and  c  are  equal,  FG  is  a  third  proportional  to  a  and  &. 


D/- 


462.    Corollary  IV.     To  divide  a  line  into  parts  propor- 
tional to  given  lines. 

Let  AB  he  the  given  line. 

Required  to  divide  it  into  parts  pro- 
portional to  the  lines  1,  2,  3,  and  4. 

Draw  AC,  making  any  convenient 
angle  with  Jfi.  Lay  off  ^D  =  1,  Z)£  =  2, 
EF  =  3,  and  FG  =  4. 

Connect  G  and  B. 

Through  F,  Ey  and  D  draw  parallels 
to  GB. 

AH  _  HI  _   IJ  _  JB 
AD~  DE~~  EF~  FG 


E/- 


F/- 


c 


Then 


or 


AH 
1 


m 

~2 


3  ~   4  ' 


Q.E.F. 


463.  Exercise.  In  the  triangle  ABC,  AB  is  10  in.  and  ^C  is  8  in. 
From  a  point  D  on  the  line  AB,  DE  is  drawn  parallel  to  BC,  making 
AD  =  3  in.     Find  the  lengths  of  AE  and  EC. 

464.  Exercise,  Through  the  point  of  intersection  of  the  medians  of 
a  triangle,  a  line  is  drawn  parallel  to  any  side  of  the 

triangle.      How  does  it  divide  each  of  the  other  two  g 

sides  of  the  triangle  ? 

Suggestion.    Use  §  245. 

465.  Exercise.  Through  a  point  within  an  an- 
gle draW  a  line  limited  by  the  sides  of  the  angle  and 
bisected  by  the  point. 

Through  the  given  point,  P,  draw  PD  \\  to  BC, 
and  lay  off  DE  =  DB. 


144 


PLANE    GEOMETBY 


466.  Exercise.  ABC  is  any  angle  and  P  a  point  within.  To  draw 
througli  P  a  line  limited  by  the  sides  of  the  angle,  and  cutting  off  a  tri- 
angle whose  area  is  a  minimum. 

Draw  HD  so  that  HP  =  PD. 
A  HBD  is  the  minimum  A. 
Draw  any  other  line  through  P,  as  EF. 
Draw  DG'  II  to  BA. 

A  PEH  =  A  PGD.  .:  A  EBF  exceeds 
area  of  A  HBD  by  A  DGF. 

467.  Exercise.  Construct  a  fourth  pro- 
portional to  three  lines  in  the  ratio  of  2,  3, 
and  4. 

468.  Exercise.  Construct  a  third  pro- 
portional to  two  lines  whose  lengths  are  1  in. 
and  3  in.  respectively. 

469.  Exercise.  Through  a  point  P  with- 
out an  angle  ABC,  draw  PE  so  that  PD—DE. 

470.  Exercise.  In  the  triangle 
ABC^  D  is  the  middle  point  of.  BC 
and  Cf  is  any  other  point  on  BC. 
Prove  that  the  parallelogram  DEAF 
is  greater  than  the  parallelogram 
GHAJ. 

Suggestion.      Draw  LK  so  that  ^ 
LG  =  GK. 

AABOAALK,     (?)  DEAF=l/\ABC,     (?) 

and  GHAJ  =  |  A  ALK.     (?) 

471.  Exercise.  Divide  a  line  into  any  number  of  equal  parts,  using 
the  principle  of  this  proposition.  Compare  the  method  with  that  used  in 
§240. 


472.   Exercise. 
proposition. 


Prove  §  239,  using  the  principle  established  in  this 


473.  Exercise.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  and 
through  the  center  of  the  circle  lines  are  drawn  parallel  to  the  sides  of  the 
triangle,  these  lines  trisect  the  sides  of  the  triangle. 


BOOK  III  -  145 

Propositiox  XIV.     Theorem  (Converse  of  Prop.  XIII.) 

474.  //  a  line  divides  two  sides  of  a  triangle  propor- 
tionally, it  is  parallel  to  the  third  side. 


Let 


AD  __AE 
DB~~  EC 

To  Prove  DE  parallel  to  BC. 

Proof.     Suppose  DE  \s  not  parallel  to  BC  and  that  any  other 
line  through  z>,  as  DM,  is  parallel  to  BC. 

AD  _AM 
DB  "  MC 


AD  _AE 
DB~  EC 


(?) 
(?) 
(?) 


AM  _AE 
MC~  EC 

Show  that  this  last  proportion  is  absurd. 
Therefore  the  supposition  that  DE  is  not  parallel  to  BC  is 
false.  Q.E.D. 

475.  Corollary.     If  —  =  — ,  DE  and  BC  are  parallel. 

A  B  A  Ky 

476.  Exercise.  DE  is  drawn,  cutting  the  sides  AB  and  AC  of  ?t 
triangle  ABC  at  D  and  E.  The  segment  BD  is  \  of  AB,  and  AE  is  f  of 
AC.     Show  that  DE  and  BC  are  parallel. 

477.  Definition.  Two  polygons  are  similar  when  they  are 
mutually  equiangular,  and  have  their  sides  about  the  equal 
angles  taken  in  the  same  order  proportional. 

SANDERS'    GEOM.  10 


146  PLANE   GEOMETRY 


Proposition  XV.     Theorem 


478.    Triangles   that  are   fnutually  equiangular  are 
uinilar. 


Let  ABC  and  DEF  be  two  A  having  /.  A  =  Z.D,  /.  B  =  /.E^ 
and  Z.C  =  /.  F. 

To  Prove  A  ABC  and  DEF  similar. 

Proof.     Lay  off  EM  =  BA,  EN  =  BC.     Draw  MN. 
Prove  A  ABC  and  DEF  equal  in  all  respects. 

Whence  /.M  =^  Z  D. 

MN  and  DF  are  II.     (?) 

EM_EN_      ,r,^  AB  ^BC 

ED~  EF  "^  DE~  EF 

T  .     -1  AB       AC  .    BC       AC 

In  a  Similar  manner  prove  —  =  — ,  and  —  = 

DE       DF  EF       DF 

The  triangles  are  by  hypothesis  mutually  equiangular,  and 
we  have  proved  their  sides  proportional,  therefore  by  definition 
they  are  similar.  q.e.d. 

479.  Corollary.  Two  triangles  are  similar  if  they  have  two 
angles  of  one  equal  respectively  to  two  angles  of  the  other. 

480.  Exercise.     All  equilateral  triangles  are  similar, 

481-  Exercise.  Are  all  isosceles  triangles  similar  ?  Are  right-angled 
isosceles  triangles  similar  ? 


BOOK  III 


147 


482.  Exercise.  If  the  sides  of  a  tri- 
angle ABC  be  cut  by  any  transversal, 
in  the  points  Z),  £',  and  F,  to  prove 


DB     EC     FA 


[From  A^  B,  and  O,  draw  perpendic-    A 
ulars    to  the   transversal.      Shovsr  that 
A  AxD  and  DyB  are  similar, 

whence  AR^Ax^ 

DB     By 


Similarly, 


and 


BE^By 

EC~  Cz' 

CF^Cz 
FA     Ax 


(1) 
(2) 
(3) 


Multiply  (1),  (2),  and  (3)  together,  member  by  member.] 

Note.  Prove  this  exercise  when  the  points  Z>,  E^  and  F  are  all 
external,  i.e.  are  all  on  the  prolonged  sides  of  the  triangle.  (If  the  figure 
be  lettered  as  above,  the  proportions  in  the  proof  of  this  case  will  be 
precisely  like  the  foregoing.) 

483.  Exercise.  If  Z>,  E  and  F  are 
three  points  on  the  sides  of  a  triangle, 
either  all  external,  or  two  internal  and 
one  external,  such  that 

AD^BEx^=\ 

DB     EC     FA       '  r.  C  G 

the  three  points  are  in  the  same  line. 

[Draw  DE  and  EF.  Let  any  other  line  than  EF  as  EG  be  the  pro- 
longation of  DE.     By  the  preceding  exercise 


DB     EC     GA 


By  hypothesis 

From  (1)  and  (2)  we  derive 


AD^BEy^QI=l 
DB     EC     FA 


CG 
GA 


FC 
FA 


0) 


(2) 


(3) 


148 


PLANE    GEOMETRY 


Arrange  (3)  by  division, 
CG 


FC         ^^   CG 

,  or         -  : 

FC         AC 


FC 
AC 


GA  -  CG     FA 
Whence  CG  =  FC  which  is  absurd. 

.  •.  the  supposition  that  any  other  line  than  EF  is  the  prolongation  of 
DE  is  absurd.] 

484.  Exercise.  If  from  any  point  on  the  cir- 
cumference of  a  circle  circumscribed  about  a  triangle 
perpendiculars  be  drawn  to  the  three  sides  of  the 
triangle,  the  feet  of  these  perpendiculars  are  in  the 
same  straight  line. 

[To  prove  x,  y,  and  z  are  in  a  straight  line. 

Connect  F  with  the  three  vertices. 

By  means  of  similar  triangles,  show  : 

Az  _Pz 
Cx  Px 
Cy^Py^ 
Bz  Pz 
Bx_Px 
Ay  Py 
Multiply  (1),  (2),  and  (3)  together,  member  by  member, 


Az^^Cy^Bx 
Cx     Bz     Ay 


Az^Bx^Cy^, 
zB      xC     yA 


By  the  preceding  exercise,  x,  y,  and  z  are  in  the  same  straight  line.  ] 

485.  Exercise.  If  a  triangle  ABC  be  inscribed  in  a  circle,  tangents 
to  this  circle  at  A,  B,  and  C  meet  BC,  CA,  and  AB  respectively  in  three 
points  that  are  in  the  same  straight  line. 

[Let  the  tangents  meet  BC,  CA,  and  AB  in  the  points  x,  y,  and  z 
respectively.     Prove  AAzC  and  BzC  similar. 


Whence 


Az_ 
AC 


zG 


(1)  and 


BC 
Bz 


AC 

Cz 


(2) 


Combining  (1)  and  (2),         ^  =  4^ 
zB      BC^ 


Similarly, 


BOOK  III  149 

Proposition  XVI.     Theorem 

486.    Triangles   that   have   their   corresponding    sides 
proportional  are  similar. 


Let  ABC  and  DBF  be  two  A  having 


AB       BC       AC 

DE       EF      -DF 

To  Prove  A  ABC  and  DEF  similar. 

Proof.     Lay  off  EM  =  BA  and  EN  =  BC. 

Draw  MN. 

Show  that                          EM^EN^ 
ED        EF 

MN  is  parallel  to  DF.     (?) 
Prove  A  EMN  and  EDF  similar. 

Whence  en^mn_ 

EF       DF  ^  ^ 

By  hypothesis  ^  =  ^.  (2) 

Compare  (1)  and  (2),  remembering  that  BC=EN,  and  show 
that  AC=  MN. 

Prove  A  ABC  and  MEN  equal  in  all  respects. 

A  DEF  and  MEN  have  been  proved  similar,  and  since  A  ABC 
and  MEN  are  equal  in  all  respects,  A  DEF  and  ABC  are 
similar.  q.e.d 


150  PLANE    GEOMETRY 

487.  Exercise.  The  sides  of  a  triangle  are  6  in.,  8  in.,  and  12  in. 
respectively.  The  sides  of  a  second  triangle  are  6  in.,  3  in.,  and  4  in. 
respectively.     Are  they  similar  ? 

488.  Scholium.  Polygons  must  fulfill  two  conditions  in 
order  to  be  similar,  i.e.  they  must  be  mutually  equiangular,  and 
must  have  their  corresponding  sides  proportional.  Proposi- 
tions XV.  and  XVI.  show  that  in  the  case  of  triangles,  either 
of  these  conditions  involves  the  other.  Hence  to  prove  triangles 
similar,  it  will  be  sufficient  to  show  either  that  they  are  mutu- 
ally equiangular,  or  that  their  corresponding 

sides  are  proportional. 

489.  Exercise.  A  piece  of  cardboard  8  in. 
square  is  cut  into  4  pieces,  A.,  B,  C,  and  2>,  as  shown 
in  the  first  figure.  These  pieces,  as  placed  in  the 
second  figure,  apparently.,  form  a  rectangle  whose 
area  is  65  sq.  in. 

Explain  the  fallacy  by  means  of  similar  triangles. 

490.  Exercise.  The  sides  of  a  triangle  are  12,  16,  and  24  ft.  re- 
spectively. A  similar  triangle  has  one  side  8  ft.  in  length.  What  is  the 
length  of  the  other  tw^o  sides  ?     (Three  solutions.) 

491.  Exercise.  On  a  given  line  as  a  side  construct  a  triangle  similar 
to  a  given  triangle.     [Construct  in  tw^o  ways.     Use  §  478  and  also  §  486.] 

492.  Exercise.  Construct  a  triangle  that  shall  have  a  given  perime- 
ter, and  shall  be  similar  to  a  given  triangle. 

493.  Exercise.  If  the  sides  of  one  triangle  are  inversely  proportional 
to  the  sides  of  a  second  triangle,  the  triangles  are  not  necessarily  similar. 

[Let  the  sides  of  the  first  triangle  be  in  the  ratio  of  2,  3,  and  4.  Then 
the  sides  of  the  second  triangle  are  in  the  ratio  of  |,  ^,  and  ^,  or  y^^,  j\, 
and  y\  ;  and  these  fractions  are  in  the  ratio  of  the  integers  6,  4,  and  3. 
Therefore  the  triangles  are  not  similar.] 

494.  Exercise.  Any.  two  altitudes  of  a  triangle  are  inversely  pro- 
portional to  the  sides  to  which  they  are  respectively  perpendicular. 


BOOK  III 


151 


Propositiox  XVII.     Theorem. 

495.    Triangles  that  have  an  angle  in  each  equal,  and 
the  including  sides  proportional,  are  similar. 


AD  and^  =  ^. 


BE      DF 


Let  A  ABC  and  BEF  have  Z  A 
To  Prove  A  ABC  and  BEF  similar. 
Proof.     JjSijoE  BM  =  AB  Sind  BN  =  AC.     Draw  J/i\r. 
Prove  A  ABC  and  BMN  equal  in  all  respects. 
BM 

BE 

MN  and  EF  are  parallel.     (?) 
Z1  =  Z2  and  Z3  =  Z4.     (?) 
A  BMN  and  BEF  are  similar.     (?) 
A  ABC  2iX\dL  BEF  are  similar.     (?) 


^.  (?) 

BF  ^ 


Q.E.D. 


496.    Exercise,     If  a  line  is  drawn  parallel  to  the  base  of  a  triangle, 
and  lines  are  drawn  from  the  vertex  to  different  points  of  the  base,  these 
lines  divide  the  base  and  the  parallel  proportionally. 
^  DBI  and  ABF  are  similar.     (?) 
DI^BI 
' '  AF     BF 
A  IB  J  and  FBG  are  similar. 
IJ  ^  BI 
"  FG     Bf' 
BI       IT 


AF     FG 


etc. 


152  PLANE    GEOMETRY 

m 

Proposition  XVIII.     Theorem. 

497.    Triangles  that  have  their  sides  parallel,  each  to 
each,  or  perpendicular,  each  to  each,  are  similar. 


Let   A  ABC  and  DEF  have  AB  II  to  DE,  BCW  to  EF,  and 

^C  II  to  DF. 

To  Prove  A  ABC  and  Z)Zi^  similar. 

Proof.  The  angles  of  the  A  ABC  are  either  equal  to  the 
angles  of  A  DEF,  or  are  their  supplements.  (§  131  and 
§  132.) 

There  are  four  possible  cases : 

1.  The  three  angles  of  A  ABC  may  be  supplements  of  the 
angles  of  A  DEF. 

2.  Two  angles  of  A  ABC  may  be  supplements  of  two  angles 
of  A  DEF,  and  the  third  angle  of  A  ABC  equal  the  third  angle 
of  A  DEF, 

3.  One  angle  of  A  ABC  may  be  the  supplement  of  an  angle 
of  A  DEF,  and  the  two  remaining  angles  oiAABC  be  equal  to 
the  two  remaining  angles  of  A  DEF. 

4.  The  three  angles  oiAABC  may  equal  the  three  angles  of 
A  DEF. 

Show  that  in  the  first  case  the  sum  of  the  angles  of  A  ABC 
would  be  four  right  angles. 

Show  that  in  the  second  case  the  sum  of  the  angles  oiAABC 
would  be  greater  than  two  right  angles. 

Show,  by  means  of  §  140,  that  the  third  case  is  impossible 


BOOK   III 


153 


unless  the  angles  that  are  supplementary  are  right  angles, 
in  which  case  they  would  also  be  equal,  and  the  triangles 
would  have  three  angles  of  the  one  equal  to  three  angles  of 
the  other. 

Therefore  if  two  triangles  have  their  sides  parallel,  each  to 
each,  the  triangles  are  mutually  equiangular,  and  consequently 
similar. 

Let  A  ABC  and  DEF  have 
ABA.de,  BC±EF,    and   AC 

_L  JJF. 

To  Prove  A  ABC  and  DEF 
similar. 

Proof.    The  angles  ofA^jSC     ^  ^ 

are  either  equal  to  the  angles  of  A  DEF,  or  are  their  supple- 
ments. 

[Show,  as  was  done  in  the  first  part  of  this  proposition,  that 
the  angles  of  A^^C  are  equal  to  those  of  A  DEF,  and  conse- 
quently A  ABC  and  DEF  are  similar.]  q.e.d. 

Note.  The  equal  angles  are  those  that  are  included  between  sides 
that  are  respectively  parallel  or  perpendicular  to  each  other. 


498.  Exercise.  The  bases  of  a  trapezoid  are  8  in.  and  12  in.,  and 
the  altitude  is  G  in.  Find  the  altitudes  of  the  two  triangles  formed  by 
producing  the  non-parallel  sides  until  they  meet. 


499.  Exercise.     The  angles  ABC,  DAE,  and 

DBE  are  right  angles. 

Prove  that  two  triangles  in  the  diagram  arc 
similar. 

500.  Exercise.     The  lines  joining  the  middle     b 
points  of  the  sides  of  a  given  triangle  form  a  sec- 
ond triangle  that  is  similar  to  the  given  triangle. 


501.    Exercise.    The  bisectors  of  the  exterior  angles  of  an  equilateral 
triangle  form  by  their  intersection  a  triangle  that  is  also  equilateral. 


154  PLANE   GEOMETRY 

Proposition  XIX.     Theorem. 

502.  The  bisector  of  an  angle  of  a  triangle  divides  the 
opposite  side  into  segments  that  are  proportional  to  the 
adjacent  sides  of  the  angle. 


Let  BD  be  tlie  bisector  of  Z  B  of  the  A  ABC, 

_     „                                             AD       AB 
To  Prove  — = 

DC       BC 

Proof.     Prolong  AB  until  BE  =  BC.     Draw  CE. 

Z3  +  Z4  =  Z1  +  Z2.     (?) 

Z  3  =  Z  4  and  Z 1  =  Z  2.     (?) 

Z  4  =  Z  2.     (?) 

BD  and  EC  are  parallel.     (?) 


BE       DC 

(?) 

AB       AD 
BC      DC 

(?) 

Q.E.D. 


Conversely.     A  line  drawn  through  the  vertex  of  an  angle 
of  a  triangle,  dividing  the  opposite  side  into  segments  pro- 


BOOK  III  155 

portional   to   the    adjacent    sides   of   the    angle,   bisects    the 
angle. 

Let  AliC  be  a  A  in  which  BD 

.     T  T  .        AD      AB 

IS  drawn  making  —  = 

DC       BC 

To  Prove  that  BD  bisects  Z.B. 


Proof.     Prolong  AB   until  BE 
=  BC.     Draw  ^'C. 


BD  is  parallel  to  EC.     (?) 
Z3  =  Z1,  and  Z4  =  Z2.     (?) 
Since  Z1  =  Z2.     (?) 

Z3  =  Z4.       (?)  Q.E.D. 

503.  Exercise.  The  triangle  ABC  has  AB  =  8  in.,  BC  =  6  in.,  and 
AC  =  12  in.  BD  bisects  ZB.  What  are  the  lengths  of  the  segments 
into  which  it  divides  AC? 

504.  Exercise.  BD  is  the  bisector  ot  Z  B  in  the  triangle  ABC. 
The  segments  ot  AC  are  ^Z>  =  5  in.  and  DC  =  2  in.  The  sum  of  the 
sides  AB  and  BC  is  14  in.     Find  the  lengths  oi  AB  and  BC. 

505.  Exercise.  Construct  a  triangle  having  given  two  sides  and  one 
of  the  two  segments  into  which  the  third  side  is  divided  by  the  bisector 
of  the  opposite  angle.     (Two  constructions.) 

506.  Definition.  A  point  C,  taken  on  the  line  AB  between 
the  points  A  and  B,  is  said  to  divide  the  line  AB  internally  into 
two   segments,  CA   and  CB. 

A  point  C',  taken  on  AB  pro-  ^  c         B — C' 

duced,  is  said  to  divide  AB 

externally  into  two  segments,  C'A  and  C'B.    In  each  case,  the  seg- 
ments are  the  distances  from  C  (or  C')  to  the  extremities  of  AB. 


156  PLANE   GEOMETRY 

Proposition  XX.     Theorem 

507.  The  bisector  of  an  exterior  angle  of  a  triangle 
divides  the  opposite  side  externally  into  two  segments 
that  are  proportional  to  the  adjacent  sides  of  the  angle. 


A 


C 


Let  BD  bisect  the  exterior  /.  CBF  oi  the  A  ABC, 

^    ^                                    AD      ab' 
To  Prove  —  = 

DC       BC 

Proof.     Lay  oR  BE  =  BC.     Draw  EC. 

Z3  +  Z4  =  Z1+Z2.     (?) 

Z3  =  Z4,  and  Z1=Z2.     (?) 

Z  4  =  Z  2.     (?) 

EC  and  BD  are  parallel.     (?) 


AB 

AD 

BE 

DC 

AB 
BC 

AD 
DC 

(?)  Q.E.D. 


508.  Exercise.  The  lengths  of  the  sides  of  a  triangle  are  4,  5,  and 
6  yards,  respectively.  Find  the  lengths  of  the  segments  into  which  the 
bisector  of  the  angle  exterior  to  the  largest  angle  of  the  triangle  divides 
the  opposite  side  externally. 


BOOK  III  157 

Conversely.  A  line  drawn  through  the  vertex  of  an  angle 
of  a  triangle  dividing  the  opposite  side  externally  into  seg- 
ments proportional  to  the  adjacent  sides  of  the  angle,  bisects 
the  exterior  angle. 

Let  BD  be  drawn  so  that  —  =  — • 

DC      BC 

To  Prove  that  BD  bisects  Z  CBF. 


O 

Proof.     Lay  of^  be  =  B  c.     Draw  CE. 

DC       BC  "^  DC       be'      ^'^ 

EC  is  parallel  to  BD.     (?) 

Z4  =  Z  1,  and  Z3  =  Z  2.     (?) 

Z  1  =  Z  2.     (?) 

Z3  =  Z4.        (?)  Q.E.D. 

509.  Definition.    A  line    Jr ^i: j,  ' 

is  divided  harmonically  when 

it  is  divided  internally  and  externally  in  the  same  ratio. 
If,  in  this  figure, 

AC _  AD 
CB  ~  DB 

then  AB  h  divided  harmonically. 

510.  Exercise.  The  bisector  of  an  angle  of  a  triangle  and  the  bisec- 
tor of  its  adjacent  exterior  angle  divide  the  opposite  side  harmonically. 
(§§  502,  507.) 


158  PLANE   GEOMETRY 

511.    Exercise.     To  divide  a  line  internally  and  externally  so  that  its 
segments  shall  have  a  given  ratio,  i.e.  to  divide  a  line  harmonically. 

Let  AB  be  the  given  line,  and  m  and  n  lines  in  the  given  ratio. 
Required  to  divide  AB  internally  and  externally  into  segments  having 

E 


n  ^^,^^  . 

Draw  AE  making  any  X    \   ^"-^      i    ^    i 

angle  with  AB.,  and  equal  /  \  ^^^^ 

to  m.  y  \  ^"-9 

Draw  BG  parallel   to         /  \  / 


^^,  and  equal  to  71.  A  G\     /-'B 

Prolong  CB  until  BD  V^- 

=  n.     Draw  ED. 

Draw  EC  and  prolong  it  until  it  meets  AB  prolonged  at  some  point  F. 

By  means  of  similar  triangles,  show 

^  =.  ^,  and  ^  =  H  ;   whence  ^  =  iZ.  q.^.p. 

GB      n  BF       n  OB     BF 

512.  Definition.  If  the  line  ^5  is  divided  harmonically 
at  G  and  D,  and  the  four  points  A,  B,  c,  and  D  are  connected 
with  any  other  point  0,  the  resulting  O 
figure  is  called  a  harmoyiic  pencil.  The 
point  0  is  called  the  vertex  of  the  pen- 
cil, and  the  four  lines  OA,  OG,  OB,  and 
OB  are  called  rays.                                     a  ^            /    j        ^P 

513.  Exercise.     0-ACBD  is   a  harmonic  pencil, 
through  G  parallel  to  OD,  and  limited 
by   OB  produced.     Prove  that  EF  is 

bisected  at  C. 

T\Tf 

(?) 

(?) 
(?) 

Multiply  (1),  (2),  and  (3)  together  member  by  member.  q.e.d. 


OD 
CF 

DB 
BG 

EG 

AG 

OD 

AD 

AG 
GB 

AD 
DB 

BOOK  III 


159 


514.    Exercise.      0-^(7-BZ>  is  a  harmonic  pencil 
versal  cutting  the  rays  at  £",   G^  II, 
and  F.     Prove  that  the  transversal 
EH  is  divided  harmonically,  that  is, 
EG  ^EF 
GH  ~  FH 
Through  C  draw  IJ  II  to  OD. 
Through  G  draw  MN  II  to  IJ. 
IC=CJ.     (?) 


.-.  MG  =  GN. 

(?) 

EG      EF      .^. 
GM      OF      ^'^ 

EG  _EF 

GN      OF 

.  EG  _EF 

"  GH     FH 

(?) 
(?) 


and  EF  any  trans- 
O 


GH 
GN 


HF 
OF 


vrxJ^   __  JLJ  ^pv 


515.  Exercise.  ABC  is  an 
inscribed  triangle,  DE  is  a  diam- 
eter perpendicular  to  AC.  The 
vertex  B  is  connected  with  the 
extremities  of  the  diameter. 
Prove  that  BE  and  DB  (pro- 
longed) divide  the  base  AC  har- 
monically. 

Suggestion.  Show  that  BE  and  BG  are  the  bisectors  of  Z  B  and  the 
exterior  angle  at  B  respectively. 

516.  Exercise.     Any  triangle  having  AC  for  its  base  (see  figure  of 

AB 

§  515),  and  its  other  two  sides  in  the  ratio ,  will  have  its  vertex  in 

BC 
the  circumference  described  on  FG  as  a  diameter, 

517.  Exercise.  The  bisectors  of  the  exterior  angles  of  a  triangle 
meet  the  opposite  sides  produced  in  three  points  that  are  in  the  same 
straight  line. 

[Let  the  bisectors  of  the  exterior  angles  at  A.,  B,  and  C,  of  the  tri- 
angle ABC,  meet  the  opposite  sides  BC,  AC,  and  AB  in  the  points  X, 
Y,  and  Z,  respectively. 

AY^AB       ,c..  CX  ^AC 

YC      BC'     ^'^  XB      AB 


(?) 


BZ 
ZA 


BC 

AC 


(?) 


Whence 


AY  ^,  CX  ^^BZ 
YC      XB      ZA 


1. 


160 


PLANE    GEOMETRY 


Proposition  XXI.     Theorem 

518.  In  a  right-angled  triangle,  if  a  perpendicular  is 
drawn  from  the  vertex  of  the  right  angle  to  the  hypote- 
nuse, 

I.  The  triangles  on  each  side  of  the  perpendicular  are 
similar  to  the  original  triangle,  and  to  each  other. 

II.   The  perpendicular  is  a  inean  proportional  between 
the  segments  of  the  hypotenuse. 

III.  Either  side  about  the  perpendicular  is  a  mean 
proportional  between  the  hypotenuse  and  the  adjacent 
segment  of  the  hypotenuse. 


Let  ABC  be  a  R.A.  A,  AC  its  hypotenuse,  and  BD  1.  to  AC. 

I.  To  Prove  Aabd  and  BBC  similar  to  A  ABC  and  to  each 
other. 

Proof.  Show  that  Aabd  and  ABC  are  mutually  equiangu- 
lar, and  consequently  similar.  In  the  same  manner  show  that 
Abdc  and  ABC  are  similar. 

Aabd  and  BDC  are  also  mutually  equiangular  and  similar. 

TT     ^     T.  ^D       BD  Q-^-^- 

II.  To  Prove  —  = 

BD       DC 

Use  the  similar  AABD  and  BDC. 


III.   To  Prove 


^  =  ^,  and   ^  =  ?^. 
AB       AD  BC      DC 


Use  the  similar  A  ABC  and  ABD,  and  also  A  ABC  and  BDC. 

Q.E.D. 


BOOK  III 


161 


519.    Corollary      To  construct  a  mean  proportio7ial  between 
two  given  lines. 

Let   m  and  n  be  two  given         ^-^ 
lines. 


■^-^.E 


Required  to  construct  a  mean     j^ 
proportional  between  them. 


B    C 


On  the  indelinite  line  AD  lay  off  A B  =  m  and  BC  =  n. 

On  ^c  as  a  diameter  describe  a  semicircle. 

Erect  BE  ±  to  AC. 

Draw  AE  and  EC. 

Show  that  A  EC  is  B.  R.A.  A,  and  that 


AB 
BE 


BE 


or  —  = 


BE 


BC  BE 

.'.  BE  is  the  required  mean  proportional 


Q.E.F. 


520.    Exercise.     Construct  a  third  proportional  to  two  given  lines  by 
means  of  Prop.  XXI. 


521.    Definition.     If  the  radius  OG 
and  externally  at  A  and  B,  so  that 

OA  X  OB  =  Og\ 


is  divided  internally 
C 


and  through  A  send  B  perpendicu- 
lars are  drawn  to  OG,  each  perpen- 
dicular is  called  the  polar  of  the 
other  point,  which  is  called  in  rela- 
tion to  the  perpendicular  its  pole. 

\^EF  is  the  polar  of  A,  and  A  is  the  pole  of  EF. 

CD  is  the  polar  of  B,  and  B  is  the  pole  of  CD. 

Notice  that  OB  is  a  third  proportional  to  OA  and  the  radius, 
and  0^  is  a  third  proportional  to  OB  and  the  radius.] 

522.    Exercise.     Given  a  point,  within  or  without  a  circle,  draw  its 
polar. 

623.    Exercise.     Given  a  line,  find  its  pole  with  respect  to  a  given 
circle. 

SANDERS'    GEOM.  11 


162 


PLANE   GEOMETRY 


524.    Exercise.     If  from  a  point  without  a  circle  two  tangents  are 
drawn  to  the  circle,  their  chord  of  contact  is  the  polar  of  the  point^ 

[To  prove  BC  the  polar  of  A. 

OA  is   ±  to   BC.      (?)      A  OB  A  is   a 
R.A.  A.     (?) 

By  Case  III.  of  this  Proposition, 


^  =  ^,  or  Oi>xO^ 
OB      OA 


0B^.-\ 


525.  Exercise.  Any  line  through  the 
pole  is  divided  harmonically  by  the  pole, 
its  polar,  and  the  circumference. 

[Let  A  be  the  pole  of  C-F,  and  EG  be 
any  line  through  A. 

To  Prove        ^  =  ^. 
AD      CD 

AO^OD     .ps       AO^OE 
OD      OB     ^"^       OE      OB 


(?) 


.*.  AAOD  and  ODB  are  similar,  as  are  also  A  OAE  and  OBE. 
AD^DB     ,p.        OE^OB     ,p.       AD^DB     ,p. 
OD      OB     ^'^       AE     EB     ^'^       AE     EB     ^' ^ 
:.  BA  bisects  Z  DBE. 

Since  CB  is  ±  to  AB,  CB  bisects  the  exterior  angle  at  B.     Now  apply 
§  510.] 

526.  Exercise,  If  two  circles  are  tan- 
gent externally,  the  portion  of  their  com- 
mon tangent  included  between  the  points 
of  contact  is  a  mean  proportional  between 
the  diameters  of  the  circles. 

[Show  that  AEB  is  a  R.A.  A,  and  that 
EF  (the  half  of  CD)  is  a  mean  proportional 
between  the  radii.] 

527.  Exercise.  If  two  tangents  are 
drawn  to  a  circle  at  the  extremities  of 
a  diameter,  the  portion  of  any  third  tan- 
gent intercepted  between  them  is  divided 
at  the  point  of  contact  into  segments 
whose  product  is  equal  to  the  square  of 
the  radius. 

fShow  that  OAB  is  a  R.A.  A.] 


BOOK  III  163 

Proposition  XXII.     Theorem 

528.  //  two  chords  intersect  within  a  circle,  the  prod- 
uct of  the  segments  of  one  is  equal  to  the  product  of 
the  segments  of  the  other. 


Let  the  chords  AB  and  CD  intersect  at  E. 
To  Prove  AE  -  EB  =  CE  •  ED. 

Proof.     Draw  AC  and  DB. 

Prove  Aaec  and  EDB  mutually  equiangular  and  therefore 
similar. 

Whence  AE^ED^  . 

CE       EB 

.-.   AE  '  EB  =  CE  '  ED.      (?)  Q.E.D. 

Conversely.  If  two  lines  AB  and  CD  intersect  at  E,  so  that 
AE  •  EB  =CE  •  ED,  then  can  a  circumference  be  passed  through 
the  four  points  A,  B,  C,  and  D. 

[Pass  a  circumference  through  the  three  points  A,  B,  and  C. 
Then  show  that  the  point  D  cannot  lie  without  this  circum- 
ference, nor  within  it.] 

529.  Exercise.  C  and  D  are  respectively  the  middle  points  of  a  chord 
AB  and  its  subtended  arc.  If  AC  is  8,  and  CD  is  4,  what  is  the  radius 
of  the  circle  ? 

530.  Exercise.  Two  chords  AB  and  CD  intersect  at  the  point  E. 
AE  is  8,  EB  is  6,  and  CD  is  19.     Find  the  segments  of  CD. 

531.  Exercise.  If  a  chord  is  drawn  through  a  fixed  point  within  a 
circle,  prove  that  the  product  of  its  segments  is  constant  in  whatever 
direction  the  chord  is  drawn. 


164 


PLANE    GEOMETRY 


Proposition  XXTII.     Theorem 

532.  //  from  a  point  without  a  circle  two  secants  he 
drawn  terminating  in  the  concave  arc,  the  product  of 
one  secant  and  its  external  segment  is  equal  to  the  prod- 
uct of  the  other  secant  and  its  external  segment. 


Let  AB  and  BC  be  two  secants  drawn  from  B  to  the  circle 
whose  center  is  0. 

To  Prove  AB  -  DB  =  CB  - EB. 

Proof.     Draw  AE  and  DC. 

Prove  A  AEB  and  CDB  mutually  equiangular  and  similar. 

BC      DB  ' 

.-.  AB  'DB  =  BC  'EB.  Q.E.D. 

Converse.  If  on  two  intersecting  lines  AB  and  CB,  four 
points,  A,  D,  C,  and  E,  be  taken,  so  that  AB  x  DB  =  BC  x  EB, 
then  can  a  circumference 
be  passed  through  the 
four  points. 

[Pass  a  circumference 
through  three  of  the 
points.  A,  D,  and  E. 
Show  by  means  of  Prop. 
XXIII.  and  the  hypoth- 
esis of  the  converse,  that  C  can  lie  neither  without  nor  within 
the  circumference.] 


BOOK  III 


165 


533.  Exercise.  One  of  two  secants  meeting  without  a  circle  is  18  in., 
and  its  external  segment  is  4  in.  long.  The  other  secant  is  divided  into 
two  equal  parts  by  the  circumference.  Find  the  length  of  the  second 
secant. 

534.  Exercise.  Two  secants  intersect  without  the  circle.  The  exter- 
nal segment  of  the  first  is  5  ft.,  and  the  internal  segment  19  ft.  long.  The 
internal  segment  of  the  second  is  7  ft.  long.  Find  the  length  of  each 
secant. 

535.  Exercise.  If  A  and  B  are  two  points  such  that  the  polar  of  A 
passes  through  B,  then  the  polar  of  B  passes  through  A. 

Let  CS^  the  polar  of  ^1,  pass 
through  B. 

To  Prove  that  the  polar  of  B  passes 
through  A. 

Proof.     [Draw  AD  ±  to  OB. 

The  quadrilateral  ADBC  has  its 
opposite  angles  supplementary,  .'.  a 
circle  can  be  circumscribed  about  it. 

ODy.  0B=  OAxOG=  6(?. 

:.  ^Z>  is  the  polar  of  B.'\ 

536.  Exercise.  The  locus  of  the  intersection  of  tangents  to  a  circle, 
at  the  extremities  of  any  chord  that  passes  through  a  given  point,  is  the 
polar  of  the  point. 

Let  CD  be  any  chord  passing  through  A,  .^^     ^VP    E 

and  B  be  the  point  of  intersection   of  the 
tangents  at  G  and  D. 

To  Prove  that  5  is  a  point  of  the  polar  of  A. 

[C2>  is  the  polar  of  B.      (§  524.) 

The  polar  of  B  therefore  passes  through  A. 
By  §  535,  the  polar  of  A  passes  through  B.  ] 

537.  Exercise.  If  from  any  point  on  a  given  line  two  tangents  are 
drawn  to  a  circle,  their  chord  of  contact  passes  through  the  pole  of  the 
line.     [Apply  §  535.] 

538.  Exercise.  If  from  different  points  on  a  given  straight  line 
pairs  of  tangents  are  drawn  to  a  circle,  their  chords  of  contact  all  pass 
through  a  common  point. 


166 


PLANE    GEOMETRY 


Proposition  XXIV.     Theorem 

539.  If  from  a  point  without  a  circle  a  secant  and  a 
tangent  are  drawn,  the  secant  terminating  in  the  con- 
cave arc,  the  square  of  the  tangent  is  equal  to  the  prod- 
uct of  the  secant  and  its  external  segment. 


Let  ^  B  be  a  tangent  and  5  C  a  secant  drawn  from  B  to  the 
circle  whose  center  is  O. 

To  Prove  AB^  =  BC  x  db. 

Proof.     Draw  AC  and  AD. 
Prove  A  CAB  and  DAB  similar. 


Whence 


BC  _AB 
AB~  DB 

-.  AB^  =  BC  X  DB.       Q.E.D. 


540.  Exercise.  Tangents  drawn  to  two 
intersecting  circles  from  a  point  on  their 
common  chord  produced,  are  equal. 

541.  Exercise.  Given  two  circles,  to 
find  a  point  such  that  the  tangents  drawn 
from  it  to  the  two  circles  are  equal. 

[Describe  any  circle  intersecting  the  two 
given  circles. 

Draw  the  two  common  chords. 

Prove  that  tangents  drawn  to  the  two 
circles  from  C,  the  point  of  intersection  of 
the  common  chords  (prolonged),  are  equal.] 


BOOK  III 


167 


Proposition  XXV.     Theorem 

542.  Two  polygons  are  similar  if  they  are  composed 
of  the  same  number  of  triangles,  similar  each  to  each, 
and  similarly  placed* 


Let  the  ^  ABC,  ADC,  DEC,  and  EFC  be  similar  respectively 
to  the  A  GHI,  GJI,  JLI,  and  LMI,  and  be  similarly  placed. 

To  Prove  polygons  ABCFED  and  GHIMLJ  similar. 

Proof.     Show  that  the  angles  of  ABCFED  are  equal  respec- 
tively to  the  corresponding  angles  of  GHIMLJ. 

AB  _AC 
GH~  Gl 


Whence 


AD 
GJ 

AB 
Gil 


Similarly  prove  —  = 


AD 
GJ 


AB 
GH 


AD 


GJ 


DE 
JL 


AC 
GI 

AD 
GJ 

DE 

EF 
LM 


(?) 
(?)' 

(?) 
etc. 

_FC  _ 
~  MI~ 


CB^ 
III 


The  polygons  are  mutually  equiangular  and  have  their 
corresponding  sides  proportional.  They  are  therefore  similar 
by  definition.  q.e.d. 


168  PLANE   GEOMETRY 

543.  Corollary.  On  a  given  line  to  construct  a  polygon 
similar  to  a  given  polygon. 

544.  Definition.  In  similar  polygons  the  corresponding 
sides  are  called  ho7nologous  sides,  and  the  equal  angles  are 
called  homologous  angles. 

Proposition  XXVI.     Theorem 

545.  Two  similar  polygons  can  be  divided  into  the 
same  number  of  simMar  triangles,  similarly  placed. 


Let  ABCDEF  and  GHIJLM  be  two  similar  polygons. 
To  Prove  that  they  can  be  divided  into  the  same  number  of 
similar  triangles,  similarly  placed. 

Proof.     From  the  vertex  F  draw  all  the  possible  diagonals. 
From  M,  homologous  with  F,  draw  all  the  possible  diagonals. 
Prove  A  FAB  and  AfGII  similar  (§  495). 
Whence  Z3=Z4. 

Z5=Z6.     (?) 

AB__B_Fl       ,c^.  A^  =  ^.      (9\ 

GH~  HM  GH       Hi' 

BFl^BC        ,^. 

HM       HI 

A  FBC  and  MHI  are  similar.    (?) 

Show  that  A  FCD  and  MIJ  are  similar,  and  also  A  FDE  and 

MJL.  Q.E.D. 


BOOK   III 


169 


Propositiox  XXVII.     Theorem 

546.   The  perimeters  of  similar  polygons  are  to  each 
other  as  any  two  homologous  sides. 

H 
C 


Let  ABODE  and  FGHIJ  be  two  similar  polygons. 
To  Prove 


AB  +  BC  +  CD  -{-  etc.  _  CD^ 
HI 


FG  4-  GH  -\-  HI  -\-  etc. 
Proof.     By  definition 

AB  _BC  _CD  _DE  _  AE 
GF  ~  GH  ~  HI  ~  IJ  ~  FJ 
[Apply  §  443.] 

547.  Corollary.  Tlie  perimeters  of  shniJar  polygons  are  to 
each  other  as  any  two  homologous  diagonals. 

548.  Exercise.  The  perimeters  of  similar  triangles  are  to  each  other 
as  any  homologous  altitudes. 

549.  Exercise.  The  perimeters  of  similar  triangles  are  to  each  other 
as  any  homologous  medians. 

550.  Exercise.  The  perimeters  of  two  similar  polygons  are  78  and 
65 ;  a  side  of  the  first  is  9,  find  the  homologous  side  of  the  second. 

551.  Definition.  A  line  is  divided  into  extreme  and  mean 
ratio  when  it  is  divided  into  two  parts  so  that  one  segment  is 
a  mean  proportional  between  the  whole  line  and  the  other 
segment. 


170 


PLANE   GEOMETRY 


Proposition  XXVIII.     Problem 
552.    To  divide  a  line  into  extreme  and  mean  ratio. 


c.--^ 


A  F  B 

Let  AB  hQ  the  given  line. 

Required  to  divide  AB  into  extreme  and  mean  ratio. 
Draw  BG  ±  to  AB  and  equal  to  one  half  of  AB.     Draw  AC. 
With  C  as  a  center  and  CB  as  a  radius  describe  a  circle 
cutting  ^C  at  Z),  and  AC  prolonged  at  E.     Lay  oE  AF  =  AD. 

£^_^^   (^8  539^       AE  —  AB      AB  —  AD 
•       AB        AD 

AD _    AB  —  AF 
AB~ 


AD 


(§  539.) 
-  (?) 


AB  AD 

AF       FB       ,ox    AB       AF 


AB       AF 


(?) 


AF       FB 


(?) 


(?)    Q.E.F. 


553.  Exercise.  To  determine  the  values  of  the  segments  of  a  line 
that  has  been  divided  into  extreme  and  mean  ratio. 

In  the  figure  of  §  552,  let  the  length  of  AB  be  a  ;  AF  =  x,  then 
FB  =  a-x. 

Substituting  these  values  in  the  last  proportion,  we  get 

-  =  — - — »   whence  a^  —  ax  =  x^. 


X     a  —  X 
Solving  the  equation, 

a;  =  ^av/5-^a=^(V5-l), 

a  _  X  =  |a  -  1  aVS  =  -  (3  -  \/5). 

554.   Exercise.     Divide  a  line  5  in.  long  into  extreme  and  mean  ratio, 
and  calculate  the  value  of  the  segments, 


BOOK  III 


171 


Proposition  XXIX.     Problem 
555.    To  draw  a  cominon  tangent  to  two  §iven  circles. 


Let  A  and  B  be  the  centers  of  the  two  given  circles. 
Required  to  draw  a  common  tangent  to  the  two  circles. 
Let  R  stand  for  the  radius  of  circle  A,  and  r  for  the  radius 
of  circle  B. 

Draw  AB.     Divide  AB  (internally  and  externally)  at  G  so 
AC 


that  ^^  = 


BC~  r 

Draw  CD  tangent  to  circle  B.     Draw  the  radius  BD. 
Draw  AE  A.  to  BC  prolonged. 
[It  is  required  to  show  that  AE  =  R.] 

Aaec  and  CBD  are  similar  (?),  whence 


AC 
BC 


AE 
BD 


J?  A  f 

—  =  — ,  and  AE  =  R,  and  ED  is  a  common  tangent, 
r        r 


Q.E.F. 


556.  Defin-ition.  The  two  tangents  that  pass  through  the 
internal  point  of  division  of  AB  are  called  the  transverse  tan- 
gents. The  two  tangents  that  pass  through  the  external  point 
of  division  are  called  the  direct  tangents. 


172 


PLANE    GEOMETRY 


The  points  of  division  are  called  the  centers  of  similitude  of 
the  two  circles. 

557.  Exercise.  The  line  joining  the  centers  of  two  circles  is  divided 
harmonically  by  the  centers  of  similitude. 

558.  Exercise,  The  line  joining  the  extremities  of  parallel  radii  of 
two  circles  passes  through  their  external  center  of  similitude  if  the  radii 
are  turned  in  the  same  direction  ;  but  through  their  internal  center  if 
they  are  turned  in  opposite  directions. 

559.  Exercise.  All  lines  passing  through  a  center  of  similitude  of 
two  circles  and  intersecting  the  circles  are  divided  by  the  circumferences 
in  the  same  ratio. 

Draw    the    radius 
AD. 

Draw   a    line    BE 
parallel   to  AD,    and 
by  means   of  similar 
triangles    prove    that 
BE  is  a  radius.    Then 
CE  ^  r 
CD      E 
Similarly, 
CE'  ^r 
CD'      B 

560.  Exercise. 
A,  Bj  and  C  are  the 

centers  of  three  cir- 
cles ;  a,  &,  and  c  their 
respective  radii  ;  D,  E, 
and  F  their  external 
centers  of  similitude  ; 
and  D',  E',  and  F' 
their  internal  centers 
of  similitude. 


Prove  that  D,  E,  and  F  are  in  a  straight  line. 
rAF^a     BD      b     .„^    CE     c      _., AF 


=z  -,    and 


whence 


BD^CE^^l 
DC     EA        J 


IFB      6'    DC     c'  EA      a 

Similarly,  show  that  D,  E',  and  F'  are  in  a  straight  line,  also  E,  D', 
and  F'  and  also  F,  D',  and  E'. 


BOOK  III  173 


EXERCISES 

1.     If        •  ^=:^ 

0      a 

-n          h  —  a     d  —  c    h  —  a      d  —  c    a      c  —  a 
Prove   = i   — : —  =  — 


c  b  d        b      d  —  b     S  a  -r  b      'S  c  +  d 

2,    If 


b~  d~ / 


prove 


xa  —  ye  +  zc  _  e^^ 
xb-yf+  zd~  f 


3.    If 


prove 


4.    If 


prove 


5.    If 


prove 


a^  +  ab  _  a 
&2  +  be  ~  c 

a  _b 
b~c' 

a^(a  +  by^ 
c      {b  +  c)^' 


6.  n  ^  =  A  :=  ^,  and  a  +  6  =  c, 

x^      2/^      ^'^ 

prove  x2  +  ?/2  —  2;2. 

7.  The  shadow  cast  by  a  church  steeple  on  level  ground  is  27  yd.,  v^hile 
that  cast  by  a  5-ft.  vertical  rod  is  3  ft.  long.     How  high  is  the  steeple  ? 

n  P 

8.  The  line   joining  the   middle   points  of  h^^    ^""A 
the  non-parallel  sides  of  a  trapezoid  circum-             /  \ 

scribed  about  a  circle  is  equal  to  one  fourth  the  /f 1\ 

perimeter  of  the  trapezoid.  /  V  y  \ 

[See  §  396.]  ^  /    ^     ^  '  \  ^ 


174 


PLANE   GEOMETRY 


9.  Two  circles  intersect  at  B  and  C. 
BA  and  BD  are  drawn  tangent  to  the 
circles. 

Prove  that  5(7  is  a  mean  proportional 
between  AG  and  CD.  [Prove  ^  ABC 
and  BCD  similar,] 

10.  Find  the  lengths  of  the  longest  and  A 
the  shortest  chords  that  can  he  drawn  through  a  point  10  in. 
center  of  a  circle  having  a  radius  26  in. 

11.  Tangents  are  drawn  to  a  circle 
at  the  extremities  of  the  diameters  AB. 
Secants  are  drawn  from  A  and  _B,  meet- 
ing the  tangents  at  D  and  E  and  inter- 
secting at  C  on  the  circumference. 

Prove  the  diameter  a  mean  propor- 
tional between  the  tangents  AD  and  BE. 
[A  ABD  and  ABC  are  similar.   (?)] 

12.  If  two  circles  are  tangent  internally,  chords  of  the  greater  drawn 
from  the  point  of  tangency  are  divided  proportionally  by  the  circum- 
ference of  the  less. 

13.  If  two  circles  are  tangent  externally,  secants  drawn  through  their 
point  of  contact  and  terminating  in  the  circumferences  are  divided  pro- 
portionally at  the  point  of  contact. 

14.  Given  the  two  segments  of  the  base  of  a  triangle  made  by  the 
bisector  of  the  vertical  triangle,  and  the  sum  of  the  other  two  sides,  to 
construct  the  triangle.     [§  502.] 

15.  Determine  a  point  P  in  the  circumference, 
from  which  chords  drawn  to  two  given  points  A  and 
B  shall  have  the  ratio  —  • 


[Divide  AB  so  that  A^  =  ^. 
*-  CB      n 

middle  point  of  the  arc  ADB.  ] 


Join  C  with  the 


16.  In  the  triangle  ABC,  BD  is  a  medial 
line,  and  DE  and  DF  bisect  angles  ADB  and 
BD  C  respectively.     Prove  that  EF  is  parallel 

to^C 


BOOK  III 


175 


17.  D  is  the  point  of  intersection  of 
the  medians  ;  E  is  the  point  of  intersection 
of  the  perpendiculars  at  the  middle  points 
of  the  sides ;  DE  is  prolonged  to  meet  the 
altitude  BI  at  F.     Prove  ED  =  l  DF. 

[A  EDG  and  DBF  are  similar,  and 
BD=2DG.'^ 


18.  The  point  of  intersection  of  the  medians,  the  point  of  intersection 
of  the  perpendiculars  at  the  middle  points  of  the  sides,  and  the  point  of 
intersection  of  the  altitudes  of  a  triangle  are  in  the  same  straight  line. 
[See  Ex.  17.] 


19.  The  triangles  ABC 
and  ADC  have  the  same 
base  and  lie  betvreen  the 
same  parallels.  EF  is  drawn 
parallel  to  AC. 

Prove  EG  =  HF. 


20.  Two  tangents  are  drawn  at  the  extremi- 
ties of  the  diameter  AB.  At  any  other  point  C 
on  the  circumference  a  third  tangent  DE  is 
drawn.  Prove  that  OD  is  a  mean  proportional 
between  AD  and  DE,  and  that  OE  is  a  mean 
proportional  between  BE  and  DE. 

[Prove  Z  DOE  a  R.A.,  and  use  §  518. J 


21.  The  prolongation  of  the  common 
chord  of  two  intersecting  circles  bisects 
their  common  tangent.     [§  539.  J 


22.  To  draw  a  line  AC  intersect- 
ing two  given  circles  so  that  the 
chords  AD  and  5C  shall  be  of  given 
lengths. 

[See  Ex.  24,  p.  126.] 


176 


PLANE   GEOMETRY 


23.  xy  is  any  line  drawn 
through  the  vertex  A  of  the 
parallelogram  ABCD  and  lying 
without  the  parallelogram. 
Prove  that  the  perpendicular  to 
xy  from  the  opposite  angle  C  is 
equal  to  the  sum  of  the  perpen- 
diculars from  B  and  D  to  xy. 
[§453.] 

24.  The  sum  of  the  perpen- 
diculars from  the  vertices  of  one 
pair  of  opposite  angles  to  a  line 
lying  without  a  parallelogram 
is  equal  to  the  sum  of  the  per- 
pendiculars from  the  vertices 
of  the  other  pair  of  opposite 
angles. 

25.  Two  circles  are  tangent  externally 
at  C.  DE  and  Ci^are  common  tangents. 
Prove  that  ZDCE=l  R.A.,  and  also 
that  Z  AFB  =  1  11.  A. 

26.  Prove  that  ^  DEC  and  CBE  (see 
figure  of  Ex.  25)  are  similar,  as  are  also 
ABAC  and  FCE.  

27.    Describe  a  circle  passing  through  two  given  points  and  tangent  to 
a  given  line. 

C  E  D  C  E  D 


v--\ 


/^ 


A\- 


'% 


(1)  (2) 

[The  line  joining  the  two  given  points  A  and  B  may  be  parallel  to  the 
given  line  CD  (see  Fig.  1),  or  its  prolongation  may  meet  the  given  line 
(see  Fig.  2).  In  the  second  case  DE'^  =  DA  x  DB.  (?)  DE  may  be 
laid  off  on  either  side  of  D,  .  •.  two  ©  can  be  described  fulfilling  the  con- 
ditions of  the  problem.] 


BOOK  III 


177 


28.  Describe  a  circle  tangent  to 
two  given  lines  and  passing  through  a 
given  point.  [P  is  the  given  point. 
Find  another  point  D  through  which 
the  circumference  must  pass.  Then 
solve  as  in  Ex.  27.] 


29.  Describe  a  circle  tangent  to 
two  given  lines  and  tangent  to  a  given 
circle.  \^DE  and  BG  are  the  lines, 
and  A  the  center  of  the  given  circle. 
Use  Ex.  28.] 


30.  Through  a  given  point  P  draw 
a  line  cutting  a  triangle  so  that  the 
sum  of  the  perpendiculars  to  the 
line  from  the  two  vertices  on  one  side 
of  the  line  shall  equal  the  perpen- 
dicular from  the  vertex  on  the  other 
side  of  the  line. 

[0  is  the  point  of  intersection  of 
the  medians.] 


31.  In  the  triangle  ABC,  BE  is  drawn 
parallel  to  AC.  FG  connects  the  middle 
points  of  ^O  and  BE.  Prove  that  FG 
prolonged  passes  through  B. 


32.  The  line  joining  the  middle  point  of 
the  lower  base  of  a  trapezoid  with  the  point 
of  intersection  of  the  diagonals  bisects  the 
upper  base. 


SANDERS'   GEOM. 


12 


178 


PLANE   GEOMETRY 


33.  In  the  triangle  ABC,  let  two  lines 
drawn  from  the  extremities  of  the  base 
AC  and  intersecting  at  any  point  D  on 
the  median  through  B,  meet  the  opposite 
sides  in  JS  and  F.  Show  that  EF  is 
parallel  to  AC. 

34.  ABC  is  an  acute-angled  triangle. 
DEF  {called  the  pedal  triangle)  is  formed 
by  joining  the  feet  of  the  altitudes  of 
triangle  ABC.  Prove  that  the  altitudes 
of  triangle  ABC  bisect  the  angles  of  the 
pedal  triangle  DEF.  [A  O  can  be  de- 
scribed passing  through  F,  0,  D,  and 
B.     (?)     Z1=Z2.     (?)] 

35.  Prove  the  triangles  AFE,  BFD, 
and  DCE  similar  to  triangle  ABC  and 
to  each  other.  [See  figure  of  Ex.  34.] 
[To  prove  i^FBD  and  ABC  similar.    Show  that  ZA  =  Z2.] 

36.  Prove  that  the  sides  of  the  triangle  ABC  [see  Ex.  34]  bisect  the 
exterior  angles  of  the  pedal  triangle  DEF. 

37.  The  three  circles  that  pass  through  two  vertices  of  a  triangle  and 
the  point  of  intersection  of  the  altitudes  are  equal  to  each  other.  [Show 
that  each  is  equal  to  the  circle  circumscribed  about  the  triangle.] 

38.  Describe  a  circle  passing  through  £^^^ -^ 

two  given  points  and  tangent  to  a  given 
circle.  [A  and  B  are  the  given  points 
and  C  the  given  circle.  DEAB  is  any 
O  passing  through  A  and  B  and  cutting 
the  given  O  C.  The  common  chord  ED 
meets  AB  at  G.  GF  is  tangent  to  O  C. 
AFB  is  the  required  O.]  G 

39.  If  one  leg  of  a  right-angled  triangle  is  double  the  other,  a  perpen- 
dicular from  the  right  angle  to    the   hypotenuse 
divides  it  into  segments  having  the  ratio  of  1  to  4. 

40.  The  triangle  ABC  is  inscribed  in  a  circle, 
and  the  bisector  of  angle  B  intersects  AC  2X  D  and 
the  circumference  at  E.     Prove 


AB 
~BD 


BE 
BC' 


BOOK  III 


179 


4i.  The  perpendicular  drawn  to  a  chord 
from  any  point  in  the  circumference  is  a 
mean  proportional  between  the  perpendicu- 
lars from  that  point  to  the  tangents  drawn 
at  the  extremities  of  the  chord. 


42.  The  perpendicular  drawn  from  the 
point  of  intersection  of  the  medians  of  a 
triangle  to  a  line  without  the  triangle  is 
equal  to  one  third  the  sum  of  the  perpen- 
diculars from  the  vertices  of  the  triangle 
to  that  line.     [§  453.] 

43.  Construct  a  right-angled  trian- 
gle, having  given  an  acute  angle  and 
the  perimeter. 

44.  Inscribe  in  a  given  triangle 
another  triangle,  the  sides  of  which 
are  parallel  to  the  sides  of  a  second 
given  triangle.  A 


45.    CD  is  a  line  perpendicular  to  the 
diameter  AB.     AE  is  drawn  from  A  to  any 
point  on    CD.     Prove   that    AE  x  AF  is  A 
constant.     [A  circle  can  be  passed  through 
F,  B,  Q,  and  E.    (?)] 


46.  Given  the  vertical  angle,  the  medial 
line  to  the  base,  and  the  angle  that  the 
medial  line  makes  with  the  base,  to  con- 
struct the  triangle. 


180 


PLANE    GEOMETRY 


47.  Given  the  base  of  a  tri- 
angle and  the  ratio  of  the  other 
two  sides,  to  find  the  locus  of  its 
vertex. 

[Divide  the  given  base  AB 
harmonically  at  D  and  E,  in  the 
ratio  of  the  two  given  sides. 
On  DE  as  a  diameter  construct 
aO.] 


48.  In  the  parallelogram  ABCD,  BE  is 
drawn  cutting  the  diagonal  AC  in  E,  CD  in  G, 
and  AB  prolonged  in  F. 

Prove  that  BE^  =  GE  x  EF. 


49.  If  three  circles  intersect  each  other,  their  common  chords  intersect 
in  the  same  point.     [§  528.] 

50.  In  any  inscribed  quadrilateral,  the  product  of  the  diagonals  is  equal 
to  the  sum  of  the  products  of  the  opposite  sides. 


61.    To  inscribe  a  square  in  a  given  semicircle. 


52.    To  inscribe  a  square  in  a  given 
triangle. 


53.  ABCD  is  a  parallelogram,  E  a  point  on  BC  such  that  BE  is  one 
fourth  of  BC.  AE  cuts  the  diagonal  BD  in  F.  Show  that  BF  is 
one  fifth  of  BD. 


54.  Two  chords  of  a  circle  drawn  from  a  common  point  A  on  the 
circumference  and  cut  by  a  line  parallel  to  a  tangent  through  A,  are 
divided  proportionally.  [Suggestion.  Join  the  extremities  of  the  chords 
and  prove  the  triangles  similar.] 


BOOK  IV 

561.  Definitions.  We  measure  a  magnitude  by  comparing 
it  with  a  similar  magnitude  that  is  taken  as  the  unit  of  meas- 
ure. If  we  wish  to  find  the  length  of  a  line,  we  find  how  many 
times  a  linear  unit  of  measure,  say  a  foot,  is  contained  in  the 
line.  This  number,  with  the  proper  denomination,  is  called 
the  length  of  the  line. 

Similarly,  we  measure  any  portion  of  a  surface  by  comparing 
it  with  some  unit  of  surface  measure.  We  find  how  many 
times  this  unit,  say  a  square  yard,  is  contained  in  the  portion 
of  surface.  This  number,  with  the  denomination  square  yards, 
we  call  the  area  or  superficial  content  of  the  surface  measured. 

Polygons  that  have  the  same  areas  are  equivalent  polygons. 
Equivalent  polygons  are  not  necessarily  equal  in  all  respects. 
They  need  not  even  have  the  same  number  of  sides.  For 
example,  a  triangle,  a  square,  and  a  hexagon  may  be  equivalent. 

The  base  of  a  polygon  is  primarily  the  side  upon  which  the 
figure  stands;  but  usage  has  sanctioned  a  more  extended  appli- 
cation of  the  term.  Any  side  of  a  polygon  may  be  considered 
the  base.  In  a  parallelogram,  if  two  opposite  sides  are  hori- 
zontal lines,  they  are  frequently  called  the  upper  and  lower 
bases  of  the  parallelogram.  In  a  trapezoid,  the  two  parallel 
sides  are  called  its  bases.  * 

The  altitude  of  a  parallelogram  is  the  perpendicular  distance 
between  two  opposite  sides.  A  parallelogram  may  therefore 
have  two  different  altitudes. 

The  altitude  of  a  trapezoid  is  the  perpendicular  distance 
between  its  bases. 

181 


182  PLANE   GEOMETRY 


Proposition  I.     Theorem 

562.  Parallelograms   having  equal    bases    and   equal 
altitudes  are  equivalent. 


B 


Let  ABCB  and  EFGH  be  two  parallelograms  having  equal 
bases  and  equal  altitudes. 

To  Prove  ^5  CD  and  EFGH  equal  in  area. 

Proof.  Place  EFGH  upon  ABCD  so  that  their  lower  bases 
shall  coincide.  Because  they  have  equal  altitudes  their  upper 
bases  are  in  the  same  line. 

Prove  A  AIB  and  DJC  equal. 

The  parallelogram  AIJD  is  composed  of  the  quadrilateral 
ABJB  and  the  Aaib. 

The  parallelogram  ABCB  is  composed  of  the  quadrilateral 

ABJB  and  the  Abjc, 

ABCB  =  AIJB. 

ABCB  =  EFGH.  Q.E.D. 

563.  Exercise.  Rectangles  having  equal  bases  and  altitudes  are 
equal  in  all  respects. 

'564.    Exercise.     Construct  a  rectangle  equivalent  to  a  given  paral- 
lelogram. 

565.  Exercise.  Prove  Prop.  I. ,  using 
this  figure  : 

566.  Exercise.  Construct  a  rectangle 
whose,  area  is  double  that  of  a  given 
equilateral  triangle. 


BOOK  IV  183 

567.  Exercise.  A  line  joining  the  middle  points  of  two  opposite 
sides  of  a  parallelogram  divides  the  figure  into  two  equivalent  parallelo- 
grams. 

Proposition  II.     Theorem 

568.  Triangles  having  equal  hases  and  equal  altitudes 
are  equivalent. 


A  C       ^ 

Let  the  ^ABC  and  BEF  have  equal  bases  and  equal  altitudes. 
To  Prove  the  i\ABC  and  BEF  equal  in  area. 
Proof.    On  each  triangle  construct  a  parallelogram  having  for 
its  base  and  altitude  the  base  and  altitude  of  the  triangle. 
These  parallelograms  are  equivalent.     (?) 
.-.  the  triangles  are  equivalent.     (?)  q.e.d. 

569.  Corollary  I.  If  a  triangle  and  a  parallelogram  have 
equal  bases  and  equal  altitudes^  the  triangle  is  equivalent  to  one 
half  the  parallelogram. 

570.  Corollary  II.  To  construct  a  triangle  equivalent  to  a 
given  polygon. 

To  construct  a  triangle  equivalent  to 
ABC  .  .  .  G. 

Draw  BB. 

Through   C  draw   CX  parallel  to  BB, 
meeting  AB  prolonged  at  X. 

Draw  BX. 

Show  that  Abxb  and  BOB  have  a  com- 
mon base  and  equal   altitudes.      .-.  Abxb  =  Abcb,  and  the 
polygon  AXBEFG  =  polygon  ABCBEFG. 


184 


PLANE   GEOMETRY 


We  have  therefore  constructed  a  polygon  equivalent  to  the 
given  polygon  and  having  one  side  less  than  the  given  polygon 
has.  A  new  polygon  may  be  constructed  equivalent  to  this 
polygon  and  having  one  side  less;  and  this  process  can  be 
repeated  until  a  triangle  is  reached. 

571.  Exercise.  Two  triangles  are  equivalent  if  they  have  two  sides 
of  the  one  equal  respectively  to  two  sides  of  the  other,  and  the  included 
angles  supplementary.  [Place  the  A  so  that  the  two  supplementary  A 
are  adjacent  and  a  side  of  one  A  coincides  with  its  equal  in  the  other.] 

572.  Exercise.     Bisect  a  triangle  by  a  line  drawn  from  a  vertex. 


573.  Exercise.  Bisect  a  triangle  by  a  line 
drawn  from  a  point  in  the  perimeter. 

[BD  is  a  medial  line,  BE  is  drawn  II  to  FD. 
Show  that  PE  bisects  AABC.^ 

574.  Exercise.  The  diagonals  of  a  paral- 
lelogram divide  it  into  four  equivalent  triangles,    a 

575.  Exercise.  The  three  medial  lines 
of  a  triangle  divide  it  into  six  equivalent  tri- 
angles. 

576.  Exercise.  In  the  triangle  ABC,  X 
is  any  point  on  the  median  CD.  Prove  that 
the  triangles  AXC  and  BXC  are  equivalent.         ^ 


577.  Exercise.  On  the  base  of  a  given  triangle  construct  a  second 
triangle  equal  in  area  to  the  first,  and  having  its  vertex  in  a  given 
straight  line.     Under  what  conditions  is  this  exercise  impossible  ? 

578.  Exercise.  Construct  a  right-angled  triangle  equivalent  to  a 
given  equilateral  triangle. 

579.  Exercise.  From  a  point  in  the  perimeter  of  a  parallelogram 
draw  a  line  that  shall  divide  the  parallelogram  into  two  equivalent  parts. 


580.    Exercise.     Construct  an  isosceles  triangle  equivalent  to  a  given 
square. 


BOOK  IV 


185 


Propositiox  III.     Theorem 

581.  Rectangles  having  equal  bases  are  to  each  other 
as  their  altitudes. 


B 


|s 


Case  I.    When  the  altitudes  are  commensurable. 

Let  ABCD  and  EFGH  be  parallelograms  having  equal  bases 
and  commensurable  altitudes. 


To  Prove 


ABCD 
EFGH 


AB 

EF 


Proof.     Let  S  be  the  unit  of  measure  for  the  altitudes,  and 
let  it  be  contained  in  AB  m  times  and  in  EF  n  times,  whence 


AB 
EF 


(1) 


Divide  the  altitudes  by  the  unit  of  measure  and  through  the 
points  of  division  draw  parallels  to  the  bases. 

ABCD  is  divided  into  m  parallelograms  and  EFGH  into  n 
parallelograms,  and  these  parallelograms  are  all  equal  by 
§  562. 

ABCD      m 


EFGH       n 


(2) 


Apply  Axiom  1  to  (1)  and  (2). 

ABCD 
EFGH  ~ 


AB 

EF 


q.e.d 


186 


PLANE   GEOMETRY 


Case  II.     When  the  altitudes  are  incommensurable. 


E  H 

Let  the  parallelograms  ABCD  and  EFGH  have  equal  bases 
and  incommensurable  altitudes. 


To  Prove 


ABCD 


AB 

'ef' 


EFGH 

Proof.  Let  EF  be  divided  into  a  number  of  equal  parts,  and 
let  one  of  these  parts  be  applied  to  AB  as  a  unit  of  measure. 

Since  AB  and  EF  are  incommensurable,  AB  will  not  contain 
the  unit  of  measure  exactly,  but  a  certain  number  of  these 
parts  will  extend  as  far  as  /,  leaving  a  remainder  IB  smaller 
than  the  unit  of  measure. 

Through  I  draw  IJ  parallel  to  the  base  AD. 

AIJD       Al 


EFGH      EF 


by  Case  I. 


By  increasing  indefinitely  the  number  of  equal  parts  into 
which  EF  is  divided,  the  divisions  will  become  smaller  and 
smaller,  and  the  remainder  IB  will  also  diminish  indefinitely. 


Now 


AIJD 


A I 


is  evidently  a  variable,  as  is  also        , 

EFGH  EF 


and  these 


variables  are  always  equal.     (Case  I.) 

The  limit  of  the  variable ■  is 

EFGH       EFGH 


AI 


AB 


The  limit  of  the  variable  ——  is 

EF      EF 


By  §  341 


ABCD 
fJFGH 


AB 
EF 


Q.E.D 


BOOK  IV 


187 


582.  Corollary.  Rectangles  having  equal  altitudes  are  to 
each  other  as  their  bases. 

583.  Exercise.  The  altitudes  of  two  rectangles  having  equal  bases 
are  12  ft.  and  16  ft.  respectively.  The  area  of  the  former  rectangle  is 
96  sq.  ft.     What  is  the  area  of  the  other  ? 

Proposition  IV.     Theorem 

584.  Ani/  two  rectangles  are  to  each  other  as  the  prod- 
ucts of  their  bases  and  altitudes. 


A  D  E  H         X 

Let  A  BCD  and  EFGH  be  any  two  rectangles. 
ABCD       AD  X  AB 


To  Prove 


EFGH      EH  X  EF 


Proof.  Construct  a  third  rectangle  XYZR,  having  a  base 
equal  to  the  base  of  ABCD  and  an  altitude  equal  to  the  altitude 
of  EFGH. 

ABCD       AB  ,os 

(?) 

(?) 

(?)  Q.E.D. 

EFGH      EF  X  EB 


XYZR 

XY 

EFGH 
XYZR 

EH 
XR 

ABCD 

AB  X  XR 

EFGH 

XY  XEH 

ABCD 

AB  X  AD 

188 


PLANE    GEOMETRY 


585.  Exercise.  The  base  and  the  altitude  of  a  certain  rectangle  are 
5  ft.  and  4  ft,  respectively.  The  base  and  the  altitude  of  a  second  rec- 
tangle are  10  ft.  and  8  ft.  respectively.     How  do  their  areas  compare  ? 

[The  student  must  not  assume  that  the  area  of  the  first  rectangle  is  20 
sq.  ft.,  as  that  has  not  yet  been  established.] 


Proposition  Y.     Theorem 

586.  The  area  of  a  rectangle  is  equal  to  the  product 
of  its  base  and  altitude. 


Let  A  BCD  be  any  rectangle. 

To  Prove  ABCD  =axb. 

Proof.     Let  the  square  U,  each  side  of  which. is  a  linear  unit, 
be  the  unit  of  measure  for  surfaces. 


ABCB      a  X  b 


(?) 


Whence 


U         1x1 
ABCD  =ab  X  U. 
or  ABCD  =  ab  X  the  surface  unit. 

or  ABCD  =  ab  surface  units. 

This  is  usually  abbreviated  into 

ABGD  =  a  xb.     (1)  Q.E.D. 


BOOK  IV 


189 


587.  Scholium.  The  meaning  to  be  attached  to  formula  (1) 
is,  that  the  number  of  surface  units  in  a  rectangle  is  the  same 
as  the  product  of  the  number  of  linear  units  in  the  base  by  the 
number  of  linear  units  in  the  altitude. 

If  the  base  is  4  ft.  and  the  altitude  3  ft.,  the  number  of 
square  feet  (surface  units)  in  the  rectangle  is  4  x  3  or  12. 
The  area  then  is  12  square  feet. 

588.  Corollary.  TJie  area  of  any  parallelogram  is  equal  to 
the  product  of  its  base  and  altitude. 

Let   ABCD  be  any  parallelogram 
and  DE  be  its  altitude. 

To  Prove  ABCD  =  ad  x  de. 

Proof.     Draw  AF  ±  to  AD,  meet- 
ing EC  prolonged  at  F. 
Prove  ADEF  a  rectangle. 

ADEF  =  ADCB.  (?) 

ADEF  =  AD   X  ED.      (?) 
ABCD  =  AD  X  ED.      (?)  Q.E.D. 


589.  Corollary.  Any  two  parallelograms  are  to  each  other 
as  the  prod^icts  of  their  bases  and  altitudes;  if  their  bases  are 
equal  the  parallelograms  are  to  each  other  as  their  altitudes;  if  the 
altitudes  are  equal  the  parallelograms  are  to  each  other  as  their 


590.    Exercise.     Construct  a  square  equivalent  to  a  given  parallelo- 
gram. 


591.    Exercise.     Construct  a   rectangle   having 
equivalent  to  a  given  parallelogram. 


a  given   base    and 


592.  Exercise.  Of  all  equivalent  parallelograms  having  a  common 
base,  the  rectangle  has  the  least  perimeter.  Of  all  equivalent  rectan- 
gles, the  square  has  the  least  perimeter. 


190 


PLANE   GEOMETRY 


Proposition  YI.     Theorem 

593.   The  area  of  a  triangle  is  one  half  the  product  of 
its  base  and  altitude. 


A  CD. 

Let  ABC  be  any  A,  and  BD  its  altitude. 
To  Prove  ABC=\ACXBD. 

Proof.     Construct  the  parallelogram  A  QBE. 
ACBE  =  AC  X  BD.      (?) 
A  ABC  =  \  AC  X  BD.      (?)  Q.E.D. 

594.  Corollary  I.  Triangles  are  to  each  other  as  the 
products  of  their  bases  and  altitudes  ;  if  their  bases  are  equal 
the  triangles  are  to  each  other  as  their  altitudes;  if  their  altitudes 
are  equal  the  triangles  are  to  each  other  as  their  bases. 

595.  Corollary  II.  TJie  area  of  a  triangle  is  one  half  the 
product  of  its  perimeter  and  the  radius  of  the  inscribed  circle. 

[Draw  radii  to  the  points  of  tan-  b 

gency.  /^S.E 

Connect  the  center  O  with  the  three 
vertices. 

Show  that  OD  is  the  altitude  of  A    

A  OB,  and  that  OE  and  OF  are  altitudes    ^ 

oi  A  BOC  and  AOC.     Call  the  radius  of  the  inscribed  circle  r. 

AAOB  =  ^AB'r.  (?) 
A  BOC  =  i  BC  -r.  (?) 
A  AOC  =  ^  AC  -r.  (?) 
A  ABC  =  J  (AB  -\-BC  -\-  CA)  r,      (?) 


Q.E.D. ] 


BOOK  IV 


191 


596.   Corollary  III.       Calling    2  s    the   perimeter    of 


triangle  ABC,  A  ABC  =  s  r,  whence   r  = 


A  ABC 


TJie  radius  of 


the  inscribed  circle  of  a  triangle  is  equal  to  the  area  of  the  triangle 
divided  by  one  half  its  perimeter. 

597.  Exercise.  The  area  of  a  rhombus  is  equal  to  one  half  the 
product  of  its  diagonals. 

598.  Exercise.      Construct  a  square  equivalent  to  a  given  triangle. 

599.  Exercise.     Construct  a  square  equivalent  to  a  given  polygon. 

600.  Exercise.  Two  triangles  having  a  common  base  are  to  each 
other  as  the  segments  into  which  the  line  joining  their  vertices  is  divided 
by  the  common  base,  or  base  produced. 

[The  A  ABC  and  A  CD  have  the  com- 
mon base  -40;  to  prove 

A  ABC  ^  BE 
ED 


AADG 

Draw  the  altitudes  BF  and  DG. 

BE^BF      ,p.  AABC^BF 

ED     DG  A  ADC     DG 


A  ABC, 
A  ADC' 


BE 
ED 


Note.    When  the  two  triangles  are  on  the  same  side  of  the  common 
base,  BD,  the  line  joining  their  vertices  is  divided  externally  at  E. 


Prove 


TIF 

=  — ,  using  these 
A ADC     DE 


601.   Definition.      Lines   that    pass    through    a  common 
point  are  called  concurrent  lines. 


192 


PLANE   GEOMETRY 


602.    Exercise.      If    three    concurrent   lines    AO,    BO,    and    CO. 
drawn  from  the  vertices  of  the  triangle  ABC, 
meet  the  opposite  sides  m  the  points  D,  E, 

and  F,  prove    M^^E     AF^  ^ 
DC     EA     FB 

[The  point  0  may  be  v^ithin  or  vyithout  the 
triangle. 


BD     AAOB 


(?) 


DC     AAOC 

AF     AAOC 


CE^ABOC 
EA 


FB 
BD 


ABOC 


AAOB 

(?) 


(?) 


CE     AF^^^ 
DC     EA     FB       ■-■ 


7?/)       C7^       A  W 

Conversely,  if  x x =  1,  to  prove  that  the  lines  AD,  BE, 

DC      EA     FB        '       ^  ' 

and  CF  are  concurrent. 

[Draw  AD  and  CF.  Call  their  point  of  in- 
tersection O.  Draw  BO.  Suppose  BO  pro- 
longed does  not  go  to  E,  but  some  other  point 
of  AC,  as  E'. 


BD^CEl^AF^^ 
DC     E'A     FB 


(?) 


DO     EA     FB 


CE'  ^  CE 
E'A     EA 


Show  that  this  last  proportion  is  absurd. 
.-.  AD,  BE,  and  Ci^are  concurrent.] 


603.    Exercise.     Show  by  means  of  the  converse  of  the  last  exercise 
that  the  following  lines  in  a  triangle  are  concurrent. 

1.  The  medial  lines. 

2.  The  bisectors  of  the  angles. 

3.  The  altitudes. 


BOOK  IV  193 

Proposition  VII.   Theorem 

604.  The  area  of  a  trapezoid  is  one  half  the  product 
of  its  altitude  and  the  sum  of  its  parallel  sides. 


A  "        n  "D 

Let  ABCB  be  a  trapezoid,  and  mn  be  its  altitude. 
To  Prove  ABCD  =  \mn{BC  +  AD). 

Proof.     Draw  the  diagonal  AC. 

Show  that  mn  is  equal  to  the  altitude  of  each  triangle 
formed.  ^ABC  =  \mn>BC.  (?) 

AACD  =  \mn'  AD.  (?) 

ABCD  =  \mn{BC -\- AD).      (?)  q.e.d. 

605.  Corollary.  The  area  of  a  trapezoid  is  equal  to  the 
product  of  the  altitude  and  the  line  joining  the  middle  points  of 
the  non-parallel  sides. 

[EF  =  ^(BC  -^  AD)    (?)      .-.    ABCD  =mn  •  EF.^ 

606.  Exercise.  In  the  figure  for  §  604  let  ^0  =  8  in.,  AD  =  12  in., 
and  mn  =  EF.     Find  the  area  of  the  trapezoid. 

607.  Exercise.     Construct  a  square  equivalent  to  a  given  trapezoid. 

608.  Exercise.  Construct  a  rectangle  equivalent  to  a  given  trape- 
zoid and  having  its  altitude  equal  to  that  of  the  trapezoid. 

609.  Exercise.  The  triangle  formed  by  joining  the  middle  point  of 
one  of  the  non-parallel  sides  of  a  trapezoid  with  the  extremities  of  the 
opposite  side  is  equivalent  to  one  half  the  trapezoid. 

610.  Exercise.  A  straight  line  joining  the  middle  points  of  the  par- 
allel sides  of  a  trapezoid  divides  it  into  tv^^o  equivalent  figures. 

SANDERS'    GEOM.  13 


194 


PLANE    GEOMETRY 


611.  Exercise.  The  area  of  a  trapezoid  is  12  sq.  ft.  The  upper  and 
lower  bases  are  7  ft.  and  5  ft.  respectively.     Find  its  altitude. 

612.  Exercise.  The  area  of  a  trapezoid  is  24  sq.  in.  The  altitude 
is  4  in. ,  and  one  of  its  parallel  sides  is  7  in.  What  is  the  other  parallel 
side  ? 

Proposition  VIII.     Theorem 

613.  Triangles  that  have  an  angle  in  one  equal  to  an 
angle  in  the  other,  are  to  each  other  as  the  products  of 
the  including  sides. 


^ 


Let 

To  Prove 


ABC  and  DEF  have  /.B  —  Ae. 
AABC      AB  '  BC 


A  DEF       BE  '  EF 
Proof.     Lay  off  BG  =  ED  and  BH  =  EF.    Draw  GH  and  AH. 


Prove 


Aabh 
Abhg 

Aabc 


a  gbh  =  a  def. 
Aabc 


BA 
BG 


(?) 


AB  .  BC 


Abhg     bg • bh 

614.  Exercise.  Prove  §613, 
using  this  pair  of  triangles. 

615.  Exercise.  The  triangle 
ABC  has  ZB  equal  to  Z^  of  tri- 
angle DEF.  The  area  of  ^JBC  is 
double  that  of  DEF.  AB  is  8  ft., 
^C  is  6  ft.,  and  DE  is  12  ft.  How 
long  is  EF? 


BOOK  IV 


195 


Proposition  IX.     Theorem 

616.  Similar  triangles  are  to  each  other  as  the  squares 
of  their  hoinologous  sides. 


Let 

To  Prove 

Proof. 


A  ABC  and  BEF  be  similar. 
A  ABC       AB^ 


A  BEF 

be' 

Zb^ 

=  Ae. 

(?) 

A  ABC 
A  BEF 

AB  '  BC 
BE  '  EF 

(?) 

AB 

BC 

(?) 

BE 

EF 

AABC 

ab' 

m 

A  BEF       JJe' 


Q.E.D. 


617.  Exercise.     Similar  triangles  are  to  each  other  as  the  squares  of 
their  homologous  altitudes. 

618.  Exercise.  In  the  triangle  ABC, 
ED  is  parallel  to  AC,  and  CD  =  \DB. 
How  do  the  areas  of  triangles  ABC  and 
BDE  compare  ? 

619.  Exercise.  The  side  of  an  equi- 
lateral triangle  is  the  radius  of  a  circle. 
The  side  of  another  equilateral  triangle  is 

the  diameter  of  the  same  circle.     How  do  the  areas  of  these  triangles 
compare  ? 


196  PLANE    GEOMETRY 

620.  Exercise.  Two  similar  triangles  have  homologous  sides  12  ft. 
and  13  ft.  respectively.  Find  the  homologous  side  of  a  similar  triangle 
equivalent  to  their  difference. 

621.  Exercise.  The  homologous  sides  of  two  similar  triangles  are 
3  ft.  and  1  ft.  respectively.     How  do  their  area's  compare  ? 

622.  Exercise.  Similar  triangles  are  to  each  other  as  the  squares  of 
any  two  homologous  medians. 

623.  Exercise.  The  base  of  a  triangle  is  32  ft.,  and  its  altitude  is 
20  ft.  What  is  the  area  of  a  triangle  cut  off  by  drawing  a  line  parallel  to 
the  base  at  a  distance  of  15  ft.  from  the  base  ? 

624.  Exercise.  A  line  is  drawn  parallel  to  the  base  of  a  triangle 
dividing  the  triangle  into  two  equivalent  portions.  In  what  ratio  does 
the  line  divide  the  other  sides  of  the  triangle  ? 

625.  Exercise.  Draw  a  line  parallel  to  the  base  of  a  triangle,  and 
cutting  off  a  triangle  that  shall  be  equivalent  to  one  third  of  the  remain- 
ing portion. 

626.  Exercise.  Equilateral  triangles  are  constructed  on  the  sides  of 
a  right-angled  triangle  as  bases.  If  one  of  the  acute  angles  of  the  right- 
angled  triangle  is  30",  how  do  the  largest  and  smallest  equilateral  triangles 
compare  in  area  ? 

627.  Exercise.  In  the  triangle  ABC,  the  altitudes  to  the  sides  AB 
and  AC  are  3  in.  and  4  in.  respectively.  Equilateral  triangles  are  con- 
structed on  the  sides  AB  and  ^O  as  bases.     Compare  their  areas. 

628.  Exercise.  The  homologous  altitudes  of  two  similar  triangles 
are  5  ft.  and  12  ft.  respectively.  Find  the  homologous  altitude  of  a  tri- 
angle similar  to  each  of  them  and  equivalent  to  their  sum. 

629.  Exercise.  Draw  a  line  parallel  to  the  base  of  a  triangle,  and 
cutting  off  a  triangle  that  is  equivalent  to  f  of  the  remaining  trapezoid. 

630.  Exercise.  Through  0,  the  point  of  intersection  of  the  altitudes 
of  the  equilateral  triangle  ABC,  lines  are  drawn  parallel  to  the  sides  AB 
and  BC  respectively  and  meeting  AG  aXx  and  y.  Compare  the  areas  of 
triangles  ABC  and  Oxy. 


BOOK   TV 


197 


Proposition  X.     Theorem 

631.  Similar  polygons  are  to  each  other  as  the  squares 
of  their  homologous  sides. 
C 


A  £ 

Let  ABODE  and  FGIIIJ  be  two  similar  polygons. 
ABODE       CD^ 


To  Prove 


FGHIJ 


HI 


Proof.     From  the  vertex  A  draw  all  the  possible  diagonals. 
From  F,  homologous  with  A,  draw  the  diagonals  in  FGHIJ. 


Similarly  prove 

Aabc 


AABC 
AFGH 

AACD 
AFHI 

AABC 
AFGH 
AACD 
A  FHI  ' 
AACD 


AC 
Flf 

¥h^ 

AACD 
A  FHI 
AADE 
AFIJ 
AADE 


AFGH       AFHI        AFIJ 
AABC  -\- AACD  +  AADE      AaCD 


(?) 
(?) 
(?) 

(?) 


A  FGH  +  A  FHI  +  A  FIJ 
AACD 
AFHI 

ABODE 
FGHIJ 


AFHI 


(?),ov 


ABODE      AACD 


^.     (?) 

HI 

g     (?) 

HI 


FGHIJ       A  FHI 


Q.E.D. 


198  PLANE    GEOMETRY 

632.  Corollary  I.  Similar  polygons  are  to  each  other  as  the 
squares  of  their  homologous  diagonals. 

633.  Corollary  II.  In  similar  polygoyis  homologous  tri- 
angles are  like  parts  of  the  polygons. 

[This  was  shown  in  the  proof  of  the  proposition.] 

634.  Exercise.  The  area  of  a  certain  polygon  is  2i  times  the  area  of 
a  similar  polygon.  A  side  of  the  first  is  3  ft.  Find  the  homologous  side 
of  the  second. 

635.  Exercise.  The  homologous  sides  of  two  similar  polygons  are 
8  in.  and  15  in.  respectively.  Find  the  homologous  side  of  a  similar 
polygon  equivalent  to  their  sum. 

636.  Exercise.  The  areas  of  two  similar  pentagons  are  18  sq.  yds. 
and  25  sq.  yds.  respectively.  A  triangle  of  the  former  pentagon  contains 
4  sq.  yds.  What  is  the  area  of  the  homologous  triangle  in  the  second 
pentagon  ? 

637.  Exercise.  If  the  triangle  ADE  [see  figure  of  §  631]  contains 
12  sq.  in.,  and  triangle  i^/J" contains  9  sq.  in.,  how  do  the  areas  of  ABODE 
and  FGHIJ  compare  ? 

638.  Exercise.  The  homologous  diagonals  of  two  similar  polygons 
are  8  in.  and  10  in.  respectively.  Find  the  homologous  diagonal  of  a 
similar  polygon  equivalent  to  their  difference. 

639.  Exercise.  Connect  C  with  w,  the  middle  point  of  AD^  and  H 
with  n,  the  middle  point  of  FI  [see  figure  of  §  631],  and  prove 

ABODE  JCm 
FGHIJ       Hn:'' 

640.  Exercise.  If  one  square*  is  double  another  square,  what  is  the 
ratio  of  their  sides  ? 

641.  Exercise.  Construct  a  hexagon  similar  to  a  given  hexagon 
and  equivalent  to  one  quarter  of  the  given  hexagon. 

« 

642.  Exercise.     Construct  a  square  equivalent  to  f  of  a  given  square. 


BOOK  IV 


199 


Proposition  XI.     Theorem 

643.  The  square  described  on  the  hypotenuse  of  a  right- 
angled  triangle  is  equivalent  to  the  sum  of  the  squares 
described  on  the  other  two  sides. 

G 


Let  ^jBC  be  a  right-angled  triangle. 
To  Prove  BO^  =  ab^  4-  ^^ 

Proof.     Describe  squares  on  the  three  sides  of  the  triangle. 
Draw  AJ  J_  to  BC,  and  prolong  it  until  it  meets  GF  at  L. 
Draw  AF  and  BD. 

Show  that  the  A  BCD  and  ACF  are  equal  by  §  30. 
Show  that  the  A  ACF  and  the  rectangle  CJLF  have  a  common 
base  and  equal  altitudes. 

Whence,  AACF  =  ^  CJLF. 

Similarly  prove  A  BCD  =  ^A  CDE. 

ACDE  =  CJLF  (?) 
In  a  similar  manner  prove  ABHI  =  BGLJ. 


ACDE  H-  ABHI  =  CJLF  +  BGLJ, 


or 


AC  -i-AB" 


bg\ 


Q.E.D. 


200 


PLANE   GEOMETRY 


644.  Note.  The  discovery  of  the  proof  of  this  proposition  is  attributed 
to  Pythagoras  (550  b.c),  and  the  proposition  is  usually  called  the  Pythago- 
rean Proposition. 

The  foregoing  proof  is  given  by  Euclid  (Book  I.,  Prop.  47). 

A  shorter  proof  follows : 

In  the  R.A.  A  ABC,  AJis  drawn  ±  to  the  hypotenuse. 


or 


By  §  518 

BC      AC 
AC      CJ 

BC     AB 
AB      BJ 

(1) 
(2) 

Whence, 

AC^=BCx  CJ. 
AB'  =  BCx  BJ 

(3) 

(4) 

Adding  (3) 

and 

(4) 

A^ 

-\-  AB^  =  BC  {CJ  -{-  BJ) 

AC^+AB^  =  B&. 

Q.E.D. 

645.  Corollary  I.  AC  =  BC  —  AB  and  AB"  =  BC  —  AC  , 
that  is,  the  square  described  on  either  side  about  the  right  angle  is 
equivalent  to  the  square  described  on  the  hypotenuse,  diminished 
by  the  square  described  on  the  other  side. 

646.  Corollary  II.  If  the  three  sides  of  a  right-angled 
triangle  are  made  homologous  sides  of  three  similar  polygons,  the 
polygon  on  the  hypotenuse  is  equivalent  to  the  sum  of  the  poly- 
gons on  the  other  two  sides. 

Let    polygons    M,   N,   and   R   be 
similar. 

M=:N-{-R. 


To  Prove 
Proof. 


AB 

AC^ 


(?) 


Whence 


iV^- 


AB    -\- AC 


M 

r' 


B^ 


(?) 


la" 


(?) 


N-[-R 
M 


AB  -{-AC^ 


BC 
AB^-\-J^=Bc\ 


:  N-\-R=M. 


Q.E.D. 


BOOK  IV  '201 

647.  Corollary  III.     TJie  squai-e  described  on  the  hypotenuse 

is  to  the  square  described  on  either  of  the  other  sides,  as  the 

hypotenuse  is  to  the  segment  of  the  hypotenuse  adjacent  to  that 

side. 

^  BG^       BC         ^    BC^       BC 

Prove  -^=9  =  —  and   — -^  = 

AC^      JC  Ib'      bj 

648.  Corollary  IV.     The  squares  described  on  the  two  sides 

about  the  right  angle  are  to  each  other  as  the  adjacent  segments 

of  the  hypotenuse. 

^  AB^      BJ 

rrove  — s  = 

AC"      JC 

In  Exercises  649-654  reference  is  made  to  the  figure  of  §  643. 

649.  Exercise.     Show  that  BI  is  parallel  to  CE. 

650.  Exercise.     The  points  H,  A,  and  D  are  in  a  straight  line, 

651.  Exercise.     AG  and  HC  are  at  right  angles,  as  are  also  AF 
and  BD. 

652.  Exercise.     If  HG,  FD,  and  IE  are  drawn,  the  three  triangles 
HBG,  FCD,  and  EAI  are  equivalent.     [Use  §  571.] 

653.  Exercise.     The  intercepts  AM  and  AN  are  equal,     [/k  BAN 
and  CAM  are  similar  to  A  BED  and  CIH  respectively.] 

654.  Exercise.     The  three  lines  J.L,  BD,  and  HC  pass  through  a 
common  point. 

[By  means  of  similar  triangles,  show  : 

fi  =  #i  W     ff=7i  (^>     and  by  Cor.  IV,  ^=^   (3). 
MB      HB  AN     AB  JC     j^(f 

Multiply  (1),  (2),  and  (3)  together,  member  by  member. 

^^  X  —  X  —  =  1.         ..  AL,  BD,  and  HC  are  concurrent.] 
MB     JC     AN  -■ 

655.  Exercise.     The  square  described  on  the  diagonal  of  a  square  is 
double  the  original  square. 

656.  Exercise.     The  diagonal  and  side  of  a  square  are  incommen- 
surable.    [See  preceding  exercise.] 


202 


PLANE    GEOMETRY 


657.  Definition.  The  projec- 
tion of  CD  on  AB  I?,  tliat  part  of 
AB  between  the  perpendiculars 
from  the  extremities  of  CD  to  AB. 

EF  is  the  projection  of  CD  on 
AB. 

MR  is  the  projection  of  MN  on 
AB. 

658.  Exercise.     The  projection  of  a 
line  upon  a  line  parallel  to  it,  is  equal  to 
tlie  line  itself.     The  projection  of  a  line  upon  another  line  to  which  it  is 
oblique  is  less  than  the  line  itself. 


Proposition  XII.     Theorem 

659.  In  any  triangle  the  square  of  a  side  opposite  an 
acute  angle  is  equivalent  to  the  sum  of  the  squares  of 
the  other  two  sides,  diminished  by  twice  the  product  of 
one  of  these  sides  and  the  projection  of  the  other  side 
upon  it. 


Let  ^5(7  be  a  A  in  which  BC  lies  opposite  an  acute  angle, 
and  ^z>  is  the  projection  of  ^^  on  ^C. 

To  Prove  BC^  ^A^ -\-AC^  —  2  AC  ^  AD. 

Proof.     In  figure  (1)     DC  =  AC  —  AD. 

In  figure  (2)  DC  =^  AD  —  AC. 

In  either  case  D&  =  AC^-\-  ad^  —  2  AC -AD. 

Dcf  +  BD^  =  Ac''-{-AD^-\-BD^-2AC'AD      (?) 

BC^  ==  IC^  +  AB^  —  2  AC  '  AD.      (?)  Q.E.D. 


BOOK  IV  203 

660.  Exercise,     Prove  this  proposition,  using  the  projection  oi  AC 
on  AB. 

661.  Exercise.      In   a  triangle  ABC,    AB  =  6  it.,  AC  =6  ft,  and 
BC  =  7  it.     Find  the  projection  oi  AC  upon  BC. 

Proposition  XIII.     Theorem 

662.  In  an  obtuse-angled  triangle  the  square  of  the 

side  opposite  the  obtuse  angle  is  equivalent  to  the  sum 

of  the  squares  of  the  other  two  sides,  increased  by  twice 

the  product  of  one  of  these  sides  and  the  projection  of 

the  other  side  upon  it. 

B 


Let  ABC  be  an  obtuse-angled  A,  and  CD  be  the  projection  of 
BC  on  ^C  (prolonged). 


To  Prove  AB^  =  BC^  +  Ilf  +  2  AC  -  CD. 

Proof.  AD  =  AC  -\-  CD. 

AD^  =AC^  -i-CD^  -\-2AC-  CD. 
AD^  -^BD'  =  Zc^-f  Cd'^  -\-  BD^  +  2AC'  CD. 

Tb^  =AC^ -{-BC^ +  2  AC  •  CD.  Q.E.D. 

663.  Corollary  I.  Tlie  right-angled  triangle  is  the  only  one 
in  which  the  square  of  one  side  is  equivalent  to  the  sum  of  the 
squares  of  the  other  two  sides. 

664.  Exercise,  The  sides  of  a  triangle  are  6,  3,  and  5.  Is  its 
greatest  angle  acute,  obtuse,  or  right  ? 


204  PLANE   GEOMETRY 

Proposition  XIV.     Theorem 

665.  In  any  triangle  the  sum  of  the  squares  of  two 
sides  is  equivalent  to  twice  the  square  of  one  half  the 
third  side  increased  by  twice  the  square  of  the  medial 
line  to  the  third  side. 


B 


Let  ABC  be  any  A  and  BD  be  a  medial  line  to  AC, 
To  Prove  AB^  H-  B(?  =  2  AD^  +  2  BD^. 

Proof.     Case  I.     When  BD  is  oblique  to  AC. 

AB^  =  AD^  4-  BD^  -\-2AD'  DE.      (?) 
BC^ —BD' -{-BC^ —  2BC  '  BE.       (?) 

AB^  4-  BC^  =  2  AD^  +  2  BD^.         (?)  Q.E.D. 

Case  II.     When  BD  is  perpendicular  to  AC. 

AB^  =  AD^  +  Bd\      (?)  BG^'^DC"  +  Bd\      (?) 

AB^  +  Bc''^2Ad''  +  2Bd\      (?)  Q.E.D. 

666.  Corollary  I.  Tlie  sum  of  the 
squares  of  the  sides  of  a  parallelogram 
is  equivalent  to  the  sum  of  the  squares 
of  the  diagonals. 

[Apply  §  Q^^  to  ^ABC  and  ADC    J 
and  add.  the  equations.] 


BOOK   IV 


205 


667.  Corollary  II.  The  sum  of  the  squares  of  the  sides  of 
any  quadrilateral  is  equivalent  to  the  sum  of  the  squares  of  the 
diagonals,  increased  by  four  times  the  square  of  the  line  joining 
the  middle  points  of  the  diagonals. 

[To  prove 

AB^ -\-BC^  +Cff  +  DA^  =  BD^ +IC^ -\-^  Mn\ 

Draw  AM  and  CM. 

Zb'  +  i^'  =  2  Zm'  +  2  Md\     (?) 

BO"  +  Cd''  =  2  cm^  +  2  md\     (?) 

2(am^  4-  cm')  =  2(2  an"  +  2  mn\    (?) 

Add  these  three  equations,  member  by  member,  and  simplify. 
(Remember  that  4  Md"^  =  Bd'^.)     (?)] 

Show  that  Cor.  I.  is  a  special  case  of  Cor.  II. 

668.  Exercise.  In  any  triangle  the  difference  of  the  squares  of  two 
sides  is  equivalent  to  the  difference  of  the  squares  of  their  projections  on 
the  third  side. 


Exercise.  In  the  diameter  of  a  circle 
two  points  A  and  B  are  taken  equally  distant  from 
the  center,  and  joined  to  any  point  P  on  the  cir- 
cumference. Show  that  AP  -\'  PB  is  constant 
for  all  positions  of  P. 

670.  Exercise.  Two  sides  and  a  diagonal  of 
a  parallelogram  are  7,  9,  and  8  respectively.  Find 
the  length  of  the  other  diagonal. 

671.  Exercise.  ABCD  is  a  rectangle,  and  P 
any  point  from  which  lines  are  drawn  to  the  four 
vertices. 

Prove      AP^  +  CP^  =  J3P^  +  DP^- 


206 


PLANE    GEOMETRY 


672.    Exercise.     If  the  side  AC  ol  the  triangle  ABC  be  divided  at  D 
so  that  mAD  =  uDC,  and  BD  be  drawn,  prove 

mAB^  +  nBC^=  mAD'^  +  nDC^  -\-(m  +  n)BD^. 

[niAB^  = 

m^AD^  +  ^^  +  2  ^2>  .  DE).     (?) 


n(^^  +  B&  -2  DC-  DE).     (?) 

mAJD^+nDC^+im-]- ti)BD^.     (?) 
Show  that  §  666  is  a  special  case  of  this  exercise.  ] 

673.  Exercise.     The  diagonals  of  a  parallelogram  are  a  ft.  and  b  ft. 

respectively,  and  one  side  is  c  ft.     Find  the  length  of  the  other  sides. 

674.  Exercise.  Inthetriangle^jB(7(seefigureof  §672),  if  JjB=9in., 
BC  =  6  in.,  AC  =  10  in.,  and  AD  =  4  in.,  find  the  length  of  BD. 

675.  Exercise.  Find  the  lengths  of  the  medians  of  a  triangle.  [In 
the  triangle  ABC  represent  the  lengths  of  the  sides  by  a,  &,  and  c.  Show 
that 

Median  to  AC  =  ^y/2  a^  +  2  c'^  -  b^ 


Median  to  BC=  iV2  b^  +  2  c^  -  a^ 


Median  to  AB=^  V2  a^  +  2  b'^  -  c2.  ] 

676.  Exercise.  In  the  triangle  ABC,  the  lengths  of  the  sides  are 
represented  by  a,  6,  and  c  (a  being  the  length  of  BC  opposite  /.A,  etc.). 
The  sum  of  the  sides  is  called  2  8. 


a  +  b  +  c  =  2  8. 


Show  that  ^-^^-^  =  S-a, 


.    g  +  6  +  c_ 
2 


/S. 


a  —  6  +  c 


=  S-b, 


«+6-c^^_^ 


BOOK   IV 


207 


Proposition  XV.     Theorem 
677.   The  area  of  the  triangle  ABC  is 


WS{s  —  a)(s  —  b){s  —  c), 
in  whiah  a,  b,  and  c  are  the  lengths  of  the  three  sides 
and  2  s  their  sum. 


Let  ABC  be  any  A. 

To  Prove       AABC=  Vs{s  —  a){S  —  b)(s  —  c). 

Proof.     Draw  tlie  altitude  bd. 

By  §  659,  a^  =  b^  +  c^-2b'AD. 


Whence 


AD  = 


52  -I-  C^  _  gg 

26 


In  the  R.A.  AABD  by  §  645, 

Bd'  =  (^      (^'  +  ^'  -  ay^4.b'c'-(b'  +  c^  -  ay 


4.W  4  62 

^  [2  6c  -  6^  -  c^  +  an  [^  6c  +  6^  +  c^  -  a'^ 


[(g  —  6  +  c)(ft  +  6  —  c)][(6  +  c  —  a)(6  +  c  +  a)] 

4  6^ 
4  /a  -|-6_-j-_c\ /6  4-  c  —  a\fa  —  6  +  c'\/'a  +  6  —  c^ 


aSfa  —  6  +  c\/a  +  6  —  c\ 

"A^2— A~-2— / 


.-.      W  =  |(s)(5-a)(S-6)(S-c). 


BD  =  ~  Vs{s  —  a){s  —  b){s  —  c). 
6 

The  area  of  A^5C  =  ^b  -  BD. 


(a) 


Area  A^^C  =  V'S(S  —  a)(S  —  6)(s  —  c). 


Q.E.D. 


208  PLANE    GEOMETRY 

678.  Corollary  I.      Tlie  area  of  an  equilateral  triangle  is 
one  fourth  the  square  of  a  side,  multiplied  by  VS. 

[In  the  formula  for  the  area  of  any  triangle,  substitute  a  for 
b  and  also  for  c.     Area  =  J  a^V3.] 

679.  Corollary  II.     The  altitude  drawn  to  the  side  b  in  tri- 


angle  ABC  is   [See    (a)    of   §677.]   -Vs(s  —  a){S  —  b){S  —  c\ 

b 

Write  the  values  of  the  altitudes  drawn  to  a  and  c  respectively. 

680.  Exercise.  Show  that  the  altitude  of  an  equilateral  triangle  is 
^aVS.     (a  =  length  of  a  side  of  the  A.) 

681.  Exercise.  The  sides  of  a  triangle  are  5,  6,  and  7.  Find  its 
area,  and  its  three  altitudes. 

682.  Exercise.  The  area  of  an  equilateral  triangle  is  25  V3.  Find  its 
side,  and  also  its  altitude. 

683.  Exercise.  The  sides  AB,  BC,  CD,  and  DA  of  a  quadrilateral 
ABCD  are  10  in.,  17  in.,  13  in.,  and  20  in.  respectively,  and  the  diagonal 
^C  is  21  in.     What  is  the  area  of  the  quadrilateral  ? 

684.  Exercise.  Two  sides  of  a  parallelogram  are  6  in.  and  7  in. 
respectively,  and  one  of  its  diagonals  is  8  in.     Find  its  area. 

685.  Exercise.  Two  diagonals  of  a  parallelogram  are  6  in.  and  8  in. 
respectively,  and  one  of  its  sides  is  5  in.  Find  its  area,  and  the  lengths 
of  its  altitudes. 

686.  Exercise.  The  parallel  sides  of  a  trapezoid  are  6  ft.  and  8  ft. 
respectively  ;  one  of  its  non-parallel  sides  is  4  ft.,  and  one  of  its  diagonals 
is  7  ft.     Find  its  area. 

687.  Exercise.  The  area  of  a  triangle  is  126  sq.  ft.,  and  two  of  its 
sides  are  20  ft.  and  21  ft.  respectively.     Find  the  third  side. 

[The  work  of  this  problem  can  be  reduced  by  using  the  formula, 
area  =^V4  h'^c^  -  (h'^  +  c'^  -  a^)^,  and  substituting  20  and  21  for  h  and  c 
respectively.] 


BOOK  IV 


209 


Proposition  XVI.     Theorem 

688.  The  area  of  a  triangle  is  equal  to  the  product  of 
its  three  sides  divided  by  four  times  the  radius  of  the 
circumscribed  circle. 


Let  ABC  be  any  A  and  let  the  lengths  of  its  sides  be  repre- 
sented by  a,  b,  and  c,  and  the  radius  of  the  circumscribed  O  be 
called  R. 

abc 


To  Prove 


Aacb 


Proof.     Draw   the   altitude  BD,  the  diameter  BE,  and  the 
chord  EC. 

AABC=\h'BD.      (?)  (1) 

Prove  Aabd  and  BEC  mutually  equiangular  and  similar, 


whence 


BD       BC 
AB      BE 

c 

in  (1). 

AABC  =  ''^'- 

4.R 

(3) 

a 
2e' 


(2) 


Q.E.D. 


689.    Corollary.     From  the  conclusion  of  the  proposition  we 


have   Aabc  = 


abc 


whence   R  = 


abc 


4/?' 

SANDE&S'    OEOAI.  —  14 


A  A  ABC 


The   radius  of   the 


210 


PLANE   GEOMETRY 


circle  circumscribed  about  a  triangle  is  equal  to  the  product  of 
the  three  sides  divided  by  four  times  the  area  of  the  triangle. 

690.    Exercise.     The  sides  of  a  triangle  are  24  ft.,  18  ft.,  and  30  ft. 
respectively.     Find  the  radius  of  the  circumscribed  circle. 


Proposition  XVII.     Problem 

691.  To  construct  a  square  equivalent  to  the  suin  of  two 
given  squares,  or  equivalent  to  the  difference  of  two  given 
squares. 


B 


a  b 

Let  A  and  B  be  two  given  squares  and  a  and  b  a  side  of  each. 


Show  that  c  =  A-\-B. 


Show  that  D  =  A  —  B. 


692.  Corollary  I.     To  construct  a  square  equivalent  to  the 
sum  of  several  given  squares. 

a,  b,  c,  and  d  are  the  sides  of  the  given 
squares. 

Zl,  Z  2,  and  Z3  are  R.A.'s. 

Show  that  Mn'  =  a'  -[- b^ -\- c^ -\-  d\ 

693.  Corollary  II.    Construct  a  square 
having  a  given  ratio  to  a  given  square. 


BOOK  IV 


211 


Let  A  be  the  given  square,  and  m  and  n  lines  having  the 
given  ratio. 

[Represent  a  side  of   the  re- 
quired square  by  x. 

ma 


Then   oiy^  =  —  a^ 
n 


X  a. 


ma 


Construct  a  line  equal  to  —  (§  461). 

n 

Call  this  line  c.     Then  x^  =  ca. 
Find  X.     (§  519.)] 

694.  Exercise.     Construct  a  square  equivalent  to  the  sum  or  differ- 
ence of  a  rectangle  and  a  square. 

[Construct  a  square  equivalent  to  the  rectangle,  and  then  proceed  as  in 
the  proposition  itself.] 

695.  Exercise.     Construct  a  square  equivalent  to  the  sum  of  the 
squares  that  have  for  sides  2,  4,  8,  12,  and  16  units  respectively. 

696.  Exercise.     If  a  =  2  in.,  construct  lines  having  the  following 

values:    aV2j  aVZ,  aVS,  aVo,  aV?,  and  aVll. 

697.  Exercise.     If  a,  b,  and  c  are  given  lines,  construct 

a2  +  3  &c  +  4  62 


and  also    x 


/3  a^b  - 
^    a  + 


2b 


2a  +  Sc 

698.  Exercise.     Construct  a  square  whose  area  shall  be  two  thirds 
of  the  area  of  a  given  square. 

699.  Exercise.     Construct  a  right-angled  triangle,  having  given  the 
hypotenuse  and  the  sum  of  the  legs. 

Let  a  be  the  given  hypotenuse  and  b  be  the  a 

sum  of  the  legs. 

[Let  X  and  y  represent  the  legs. 
Then  jc  +  y  =  6, 

x^  +  y2  =  a^.     (§643.) 
Solving  these  equations,  we  get 

x  =  l(b  +  V2  a2  -  62), 


y=l(b-  V2  a2  _  62). 

Construct  these  values  of  x  and  y. 

Then  the  three  sides  of  the  triangle  are  known.  ] 


212  PLANE   GEOMETRY 

700.  Exercise.     Construct  a  right-angled  triangle,  having  given  one 
leg,  and  the  sum  of  the  hypotenuse  and  the  other  leg. 

Proposition  XVIII.     Problem 

701.  To  construct  a  polygon  similar  to  either  of  two 
given  similar  polygons  and  equivalent  to  their  sum. 


b 

Let  A  and  B  be  the  two  given  similar  polygons. 
Required  to  construct  a  third  polygon  similar  to  either  A  or 

B,  and  equivalent  to  their  sum. 

[Construct  a  R.A.  A  having  a  and  b  (homologous  sides  of  A 

and  B)  for  legs.      On  the  hypotenuse  of  this  A  construct  a 

polygon  similar  to  A.    Show,  by  §  646,  that  this  is  the  required 

polygon.] 

702.  Corollary  I.     Construct  a  polygon  similar  to  either  of 
two  given  ^similar  polygons  and  equivalent  to  their  difference. 

703.  Corollary  II.     Construct  a  polygon  similar  to  a  given 
polygon  and  having  a  given  ratio  to  it. 

Let  a  be  a  side  of  the  given 
polygon  A  and  —  be  the  given 
ratio. 

[Construct  oi^  =  —  a^. 

''    (§693.) 

On  a  side  of  the  square  x^  construct  a  polygon  R  similar  to  A. 

l^a^ 
R  _^  _n _m       .^.- 


r? 


BOOK  IV  213 

704.  Corollary   III.      Construct  a  polygon  similar  to  one 
given  polygon  and  equivalent  to  another. 

Let  A  and  B  be  the  two 
given  polygons. 

Required  to  construct  a 
polygon  similar  to  A  and 
equivalent  to  B. 

[Construct     a     square     C 
equivalent  to  A,  and  a  square  D  equivalent  to  B.     Let  c  and  d 
be  sides  of  these  squares. 

Construct  a  line  m  =  - — .     (§  461.) 

On  m,  homologous  with  a,  construct  a  polygon  i?  similar 
to  A.  a^ 

'     Since  A  =  c^, 

R  =  d'  =  s.] 

705.  Exercise.     Construct  a  quadrilateral  similar  to  a  given  quadri- 
lateral and  whose  area  shall  be  3  sq.  in.     (§  704.) 

706.  Exercise.     Construct  an  equilateral  triangle  the  area  of  which 
shall  be  three  fourths  of  that  of  a  given  square. 

EXERCISES 

1.  The  diagonal  of  a  rectangle  is  13  ft.,  one  of  its  sides  is  12  ft.* 
What  is  its  area  ? 

2.  The  square  on  the   hypotenuse  of    an   isosceles  right-angled   tri- 
angle is  four  times  the  area  of  the  triangle. 

3.  The  base  of  an  isosceles  triangle  is  14  in., 
and  one  of  the  other  sides  is  18  in.  Find  the 
lengths  of  its  altitudes. 

4.  Find  a  point  within  a  triangle  such  that  lines 
drawn  from  it  to  the  three  vertices  divide  the  tri- 
angle into  three  equal  parts. 


214 


PLANE   GEOMETRY 


6.  If  a  circle  is  inscribed  in  a  tri- 
angle, tlie  lines  joining  the  points  of 
tangency  with  the  opposite  vertices 
are  concurrent. 


[Showthat4?x:??x^^ 


FB     DC     EA 


1. 


6.  Given  a  triangle,  to  construct 
an  equivalent  parallelogram  the  perim- 
eter of  which  shall  equal  that  of  the 
triangle. 

[FE=l{AC-^  BC).'] 


7.  The  sum  of  the  three  perpendiculars  from  a  point  within  an  equi- 
lateral triangle  to  the  three  sides  is  equal  to  the  altitude  of  the  triangle. 

8.  The  bases  of  two  equivalent  triangles  are  10  ft.  and  15  ft.  respec- 
tively.    Find  the  ratio  of  their  altitudes.  _ 

9.  ABCD    is   any  parallelogram, 
and  0  is  any  point  within. 

Prove  that  the  sum  of  the  areas  of 
triangles  OAB  and  OCD  equals  one 
half  the  area  of  the  parallelogram. 


10.    ABC  is  a  right-angled  triangle,  and 
JSD  bisects  ^C. 

Prove  that  BD^  =  BC^  -  3  DG^. 


11.  In  the  right-angled  triangle  ABC, 
AD  is  perpendicular  to  the  hypotenuse 
BC,  and  the  segments  BD  and  DC  are  9  ft. 
and  16  ft.  respectively.  Find  the  lengths 
of  the  sides,  the  area  of  the  triangle,  and 
the  length  of  AD. 

12.  A  square  is  greater  than  any  other  rectangle  inscribed  in  the  same 
circle. 

[Show  that  both  square  and  rectangle  have  diameters  for  diagonals.  ] 


BOOK  IV 


215 


13.  ABCD  is  any  quadrilateral,  and  AE 
and  CF  are  drawn  to  the  middle  points  of 
BC  and  AD  respectively. 

Prove  AECF  equivalent  to  BEA+  CFD. 

14.  From  any  point  0  within  the  tri- 
angle ABC,  OX,  or,  and  OZ  are  drawn 
perpendicular  to  BC,  CA,  and  AB  re- 
spectively. 

'  Prove 

AZ^  +  BX^  +  CY^  =  ZB^  +  XC^  +  T^. 
[Draw    OA,    OB,   and    00.     Then    use 
§643.] 

15.  In  the  parallelogram  ABCD  any 
point  on  the  diagonal  ^O  is  joined  with  the 
vertices  B  and  D. 

Prove  triangles  ABE  and  AED  equiva- 
lent. 

16.  Draw  a  line  through  the  point  of  intersection  of  the  diagonals  of  a 
trapezoid  dividing  it  into  two  equivalent  trapezoids. 


17.  The  square  described  on  the  sum  of  two  lines 
is  equivalent  to  the  sum  of  the  squares  of  the  lines 
increased  by  twice  their  rectangle. 

18.  The  square  described  on  the  difference  of  two 
lines  is  equivalent  to  the  sum  of  the  squares  of  the 
lines  diminished  by  twice  their  rectangle. 


19.  The  rectangle  having  for  its  sides  the  sum  and 
the  difference  of  two  lines  is  equivalent  to  the  differ- 
ence of  their  squares. 


20.  A  triangle  and  a  rectangle  having  equal  bases  are  equivalent. 
How  do  their  altitudes  compare  ? 

21.  Draw  a  straight  line  through  a  vertex  of  a  triangle  dividing  it  into 
two  parts  having  the  ratio  of  w  to  n. 


216 


PLANE   GEOMETRY 


22.  Through  a  given  point  within  or  without  a  parallelogram  draw  a 
line  dividing  the  parallelogram  into  two  equivalent  parts. 

23.  If  a  and  h  are  the  sides  of  a  triangle,  show  that  its  area  =  \ah 
when  the  included  angle  is  30°  or  150°;  \ah-\/2  when  the  included  angle 
is  45°  or  135°  ;  i  a&  V3  when  the  included  angle  is  60°  or  120°. 

[Using  either  a  or  6  for  base,  find  the  altitude  of  the  A.] 

24.  If  equilateral  triangles  are  described  on  the  three  sides  of  a  right- 
angled  triangle,  prove  that  the  triangle  on  the  hypotenuse  is  equivalent 
to  the  sum  of  the  triangles  on  the  other  sides. 

25.  On  a  given  line  as  a  base  construct  a  rectangle  equivalent  to  a 
given  rhombus. 

26.  Bisect  a  triangle  by  a  line  drawn  parallel  to  one  of  its  sides. 
[§616.] 

27.  The  square  of  a  line  from  the  vertex  of 
an  isosceles  triangle  to  the  base  is  equivalent 
to  the  square  of  one  of  the  equal  sides  dimin- 
ished by  the  rectangle  of  the  segments  of  the 
base  [i.e.  B&  =  AJE^  -  AD  x  DC].  [Draw 
the  altitude  to  AC.     Use  §  643.] 

28.  If,   in  Exercise  27,  BD  is  drawn  to  a 

point  D   on  the  prolonged  base,  then  BD^  =  AB^  +  AD  x  DC. 

29.  Three  times  the  sum  of  the  squares  on  the  sides  of  a  triangle 
is  equivalent  to  four  times  the  sum  of  the  squares  on  its  medians. 
[§665.] 

30.  If  the  base  a  of  a  triangle  is  increased  d  inches,  how  much  must 
the  altitude  b  be  diminished  in  order  that  the  area  of  the  triangle  shall 
be  unaltered. 

31.  OC  is  a  line  drawn  from  the  center  of 
the  circle  to  any  point  of  the  chord  AB. 

Prove  that  OC^  =  OA^  -  AC  x  CB. 

32.  The  lengths  of  the  parallel  sides  of  a 
trapezoid  are  a  ft.  and  h  ft.  respectively.  The 
two  inclined  sides  are  each  c  ft.  Find  the  area 
of  the  trapezoid.  B 

33.  From  the  middle  point  D  of  the  base  of 
the  right-angled  triangle  ABC,  DE  is  drawn 
perpendicular  to  the  hypotenuse  BC. 

Prove  that  BE"^  -  EG^  =  AB^. 


BOOK  IV 


217 


34.  In  any  circle  the  sum  of  the  squares  on  the  segments  of  two  chords 
that  are  perpendicular  to  each  other  is  equivalent  to  the  square  on  the 
diameter.     [§  64;].] 

35.  Construct  a  triangle  having  given  its 
angles  and  its  area. 

36.  In  the  triangle  ABC,  AD,  BE,  and  GF 
are  lines  drawn  from  the  vertices  and  passing 
through  a  common  point  0. 

Prove  that  ^  +  ^+^=1. 


VOE 

Vbe 


BE 

J\AOC 


AD      CF 


(?) 


A  ABC 

37.  From  any  point  0  within  a  triangle 
ABC,  OD,  OE,  and  OF  are  drawn  to  the 
three  sides.  From  the  vertices  AD',  BE',  and 
CF'  are  drawn  parallel  to  OD,  OE,  and  OF 
respectively. 


Find  similar  expressions  for  -^—  and 


Prove  that 

OE 
BE' 


OD 
AD' 


OF 

CF' 


1. 


rOE 
\_BE' 


A 

/\AOC 


E      E' 


/\ABC  J 

38.  Given  the  altitude,  one  of   the  angles,  and  the  area,  construct 
a  parallelogram. 

39.  The  two  medial  lines  AE  and  CD  of 
the  triangle  ABC  intersect  at  F.  Prove  the 
triangle  AFC  equivalent  to  the  quadrilateral 
BDFE. 

40.  The  diagonals  of  a  trapezoid  divide  it 
into  four  triangles,  two  of  which  are  similar, 
and  the  other  two  equivalent. 

41.  Any  two  points,  C  and  D,  in  the  semi- 
circumference  ACB  are  joined  with  the  ex- 
tremities of  the  diameter  AB.  AE  and  BF 
are  drawn  perpendicular  to  the  chord  DC  pro- 
longed. 


Prove  that  CE'^  +  CF'^  =  DE'^  +  DF'^. 


A  B 

[Use  §  643.] 

42.    Describe   four  circles   each   of  which  is   tangent  to  three  lines 
that  form  a  triangle. 


218 


PLANE   GEOMETRY 


[One  of  the  four  is  the  inscribed  circle  of  the  A,  and  its  radius  is 
denoted  by  r.  The  other  three  are 
called  escribed  circles  of  the  tri- 
angle, and  their  radii  are  denoted 
hy  ^aj  Ti^  and  re.  (?*«  is  the  radius 
of  the  escribed  circle  lying  between 
the  sides  of  Z  J.  of  the  A.)] 

43.  The  area  of  triangle  ABC 
=  fa  (S  —  a). 

[A  ABC  =  A  ABE  +  A  ACE  - 
A  BEC,  and  ra  is  the  altitude  of 
each  of  these  A.] 

Show  that  rj  (S—b)  and  re  iS—c) 
are  also  expressions  for  the  area  of  triangle  ABC. 

44.  The  area  of  triangle  ABC 


Vr 


X  7'ft  X  re.     [Ex.  43.] 


45.    Prove  that  ra-\-  n  +  re —r  =  4  B  [B  =  radius  of  the  circle  cir- 
cumscribed about  A  ABC].     [Ex.  43  and  §  689.] 


46.   Prove  that 


r       Va       Th       Tc 


47.  The  area  of  a  quadrilateral  is 
equivalent  to  that  of  a  triangle  having 
two  of  its  sides  equal  to  the  diagonals 
of  the  quadrilateral  and  its  included 
angle  equal  to  either  of  the  angles  be- 
tween the  diagonals  of  the  quadrilateral. 
IDF  =  BE  and  CQ  =  AE.  Show 
that  A  GDF  =AABC  and  A  GED 
=  A  ACD.] 

48.  Parallelograms  A  DEB  and 
BFGC  are  described  on  two  sides 
of  the  triangle  ABC.  DE  and  GF 
are  prolonged  until  they  meet  at  H. 
HB  is  drawn,  A  third  parallelogram 
AIJC  is  constructed  on  AC,  having 
AI  equal  to  and  parallel  to  BH.  Prove 
that  AIJC  is  equivalent  to  the  sum  of 
ADEB  and  BFGG.  IADEB=ALKI 
and  BFGC  =  LCJK.'\ 


BOOK  IV 


219 


49.  The  lines  joining  the  points  of  tangency  of  the  escribed  circles 
with  the  opposite  vertices  of  the  triangle  ABC,  are  concurrent.   [See  Ex.  5.] 

50.  Deduce  the  Pythagorean  Theo- 
rem (Prop.  XI,  Bk.  IV)  from  Exercise 
48. 

51.  Through  a  point  P  within  an 
angle  draw  a  line  such  that  it  and 
the  parts  of  the  sides  that  are  inter- 
cepted shall  contain  a  given  area. 

[Construct  parallelogram  BDEF  = 
required  area  (Ex.  38),  DE  passing  through  P.  If  HG  is  the  required 
line,  A  PIE  =  A  IFH  +  A  PDG.  The  A  are  similar,  DP,  PE,  and  FH 
are  homologous  sides,  and  DP  and  PE  are  known.] 


i.   Is  there  any  limit  to  the  "  given  area  "  in  Exercise  51  ? 


BOOK   V 


707.  Definition.  A  regular  polygon  is  a  polygon  that  is 
both  equilateral  and  equiangular. 

Proposition  I.     Theorem 

708.  //  the  circumference  of  a  circle  is  divided  into 
three  or  more  equal  parts,  the  chords  joining  the  succes- 
sive points  of  division  form  a  regular  inscribed  polygon; 
and  tangents  drawn  at  the  points  of  division  form  a 
regular  circumscribed  polygon. 


Let  the  arcs  AB,  BC,  etc.,  be  equal. 

To  Prove  the  polygon  ABCD  •••  a  regular  inscribed  polygon. 
[The  proof  is  left  to  the  student.] 

Let  the  arcs  AB,  BC,  etc.,  be  equal. 

To  Prove  the  polygon  xyz  •••a  regular  circumscribed  polygon. 

Proof.     [Draw  the  chords  AB,  BC,  etc. 

Show  that  the  A  AyB,  BzC,  etc.,  are  isosceles  A  and  are  equal 
in  all  respects.] 

220 


BOOK   V 


221 


709.    Corollary  I.     If  at  the  middle  points  of  the  arcs  sub- 
tended by  the  sides  of  a  regular  inscribed  poly- 
gon, tangents  to  the  circle  are  drawn, 

L     The  circumscribed  polygon  formed  is 
regular. 

II.  Its  sides  are  parallel  to  the  sides  of 
the  inscribed  polygon. 

III.  A  line  connecting  the  center  of  the 

circle  with  a  vertex  of  the  outer  polygon   passes   through  a 
vertex  of  the  inner  polygon. 

[?/0  bisects  /-HON,  consequently  bisects  arc  MN,  and  there- 
fore passes  through  bJ] 


710.  Corollary  II.  If  the  arcs  subtended  by  the  sides  of  a 
regular  inscribed  polygon  are  bisected,  and  the  points  of  division 
are  joined  with  the  extremities  of  the  arcs,  the  polygon  formed  is  a 
regular  inscribed  polygon  of  double  the  number  of  sides;  and  if 
at  the  extremities  of  the  arcs  and  at  their  middle  jioiids  tangents 
are  drawn,  the  polygoyi  formed  is  a  regular  circumscribed  polygon 
of  double  the  number  of  sides. 


711.  Corollary  III.  TJie  area  of  a  regular  inscribed  poly- 
gon is  less  than  that  of  a  regular  inscribed  polygon  of  double  the 
number  of  sides;  but  the  area  of  a  regular  circumscribed  polygon 
is  greater  than  that  of  a  regular  circumscribed  polygon  of  double 
the  number  of  sides. 

712.  Exercise,  An  equiangular  polygon  circumscribed  about  a  circle 
is  regular. 

713.  Exercise.  An  inscribed  equiangular  polygon  is  regular  if  the 
number  of  its  sides  is  odd. 


714.   Exercise.     A  circumscribed  equilateral  polygon  is  regular  if  the 
number  of  its  sides  is  odd. 


222  PLANE    GEOMETRY 


Proposition  II.      Theorem 


715.  A  circle  can  he  circumscribed  about  any  regular 
polygon;  and  one  can  also  be  inscribed  in  it. 


Let  ABC  "•  GhQ2i  regular  polygon. 

I.  To  Prove  that  a  circle  can  be  circumscribed  about  it. 

Proof.  Pass  a  circumference  through  three  of  the  vertices, 
A,  B,  and  C,  and  let  O  be  its  center. 

Draw  the  radii  OA,  OB,  and  OC.    Draw  OB. 

ShowthatZl  =  iZ5andZ3  =  iZc. 

Prove  AOCB  and  OCB  equal  in  all  respects. 

Whence  OD  =  OB. 

Therefore  the  circumference  that  passes  through  A^  B,  and  C 
will  also  pass  through  D. 

Similarly,  it  can  be  shown  that  this  circumference  passes 
through  the  remaining  vertices.  q.e.d. 

II.  To  Prove  that  a  circle  can  be  inscribed  in  the  polygon. 

Proof.    Describe  a  circle  about  the  regular  polygon  AB  -"  G. 

The  sides  AB,  BC,  etc.,  are  all  equal  chords  of  this  circle,  and 
are  equally  distant  from  the  center  (?). 

With  O  as  a  center  and  this  distance  for  a  radius  describe 
a  circle. 

Show  that  AB,  BC,  etc.,  are  tangent  to  this  circle,  which  is, 
therefore,  a  circle  inscribed  in  the  regular  polygon.  q.e.d. 


BOOK   V  223 

716.  Definitions.  The  common  center  of  the  circles  that 
are  inscribed  in  and  circumscribed  about  a  regular  polygon,  is 
called  the  ceriter  of  the  x>olygon.  The  angles  formed  by  radii 
drawn  from  this  center  to  the  vertices  of  the  polygon  are  called 
angles  at  the  center.  Each  angle  at  the  center  is  equal  to  4  right 
angles  divided  by  the  number  of  sides  in  the  polygon.  A  line 
drawn  from  the  center  of  the  polygon  perpendicular  to  a  side, 
is  an  apothem.  The  apothem  of  a  regular  polygon  is  equal  to 
the  radius  of  the  inscribed  circle. 

717.  Exercise.  How  many  degrees  in  the  angle  at  the  center  of  an 
equilateral  triangle  ?  Of  a  square  ?  Of  a  regular  hexagon  ?  Of  a  regular 
polygon  of  n  sides  ? 

718.  Exercise.  How  many  sides  has  the  polygon  whose  angle  at  the 
center  is  30°  ?     18"  ? 

719.  Exercise.  In  what  regular  polygon  is  the  apothem  one  half  the 
radius  of  the  circumscribed  circle  ? 

720.  Exercise.  In  what  regular  polygon  is  the  apothem  one  half  the 
side  of  the  polygon  ? 

721.  Exercise.  Show  that  an  angle  at  the  center  of  any  regular 
polygon  is  equal  to  an  exterior  angle  of  the  polygon. 

Proposition  III.     Theorem 

722.  Regular  polygons  of  the  same  number  of  sides  are 
similar. 


[Show  that  the  polygons  are  mutually  equiangular  and  have 
their  homologous  sides  proportional.] 


224 


PLANE   GEOMETRY 


Proposition  IV.     Theorem 

723.  The  perimeters  of  similar  regular  polygons  are  to 
each  other  as  the  radii  of  their  inscribed  or  of  their  cir- 
cumscribed circles;  and  the  polygons  are  to  each  other 
as  the  squares  of  the  radii. 

N 


Let  ABC  '••  F  and  MNR  •••  -S  be  two  similar  regular  polygons. 

To  Prove  that  their  perimeters  are  proportional  to  the  radii 
of  the  inscribed  and  of  the  circumscribed  circles,  and  that  their 
areas  are  proportional  to  the  squares  of  these  radii. 

Proof.     Let  x  and  y  be  the  centers  of  the  regular  polygons. 
Draw  xB  and  yN,  and  the  apothems  xE  and  yL. 
xB  and  yN   are  the  radii  of  the  circumscribed  circles  and 
xE  and  yL  are  the  radii  of  the  inscribed  circles. 


Perimeter  ABC  ••-  F_  BC^ _Bx  _xE 
S~  NR~~  Ny~  yL 


Perimeter  MNR 
Area  ABC  -"  F 


(?) 


Area  J/iV^i?---  S 


BC 
NR^ 


Bx 
Ny 


2  2  ■ -2  ^'  ^ 


xE 

■ -') 

yL- 


Q.E.D. 


724.  Exercise.  Two  squares  are  inscribed  in  circles,  the  diameters 
of  wliich  are  2  in.  and  6  in.  respectively.     Compare  their  areas. 

725.  Exercise.  A  regular  polygon,  the  side  of  which  is  6  in.,  is 
circumscribed  about  a  circle  having  a  radius  \/3  in.  Find  the  side  of  a 
similar  polygon  circumscribed  about  a  circle  the  radius  of  which  is  6  in. 


BOOK    V 


225 


726.  Exercise.  The  perimeters  of  similar  regular  polygons  are  to 
each  other  as  the  diameters  of  their  inscribed  or  of  their  circumscribed 
circles  ;  and  the  polygons  are  to  each  other  as  the  squares  of  the  diameters. 


Proposition   V.     Problem 
727.  To  inscribe  a  square  in  a  given  circle, 

C 


Let  0  be  the  center  of  the  given  circle. 

Required  to  inscribe  a  square  in  the  circle. 
Draw  the  diameters  AB  and  CD  at  right  angles. 
Connect  their  extremities. 
Prove  ACBD  an  inscribed  square.     (§  708.) 


Q.E.F. 


728.  Corollary  I.  Tangents  to  the  circle  at  the  extremities 
of  the  diameters  AB  and  CD  form  a  circumscribed  square. 

729.  Corollary  II.     The  side  of  the  inscribed  square  is  R  ^2. 
The  side  of  the  circumscribed  square  is  2r. 

The  area  of  the  inscribed  square  is  2r^. 
The  area  of  the  circumscribed  square  is  4i2^. 

730.  Corollary  III.  By  bisecting  the  arcs  and  drawing 
chords  and  tangents  as  described  in  §  710,  regidar  polygons  of 
8,  16,  32,  64,  etc.,  sides  can  be  inscribed  in  and  circumscribed 
about  the  circle. 

SANDERS'  GEOM.  —  15 


226 


PLANE   GEOMETRY 


731.  ExKRCiSE.     The  radius  of  a  circle  is  5  ft.     Find  the  side  and  the 
area  of  the  inscribed  square. 

732.  Exercise.     Find  the  side  and  the  area  of  a  square  circumscribed 
about  a  circle,  having  a  diameter  6  in.  long. 

733.  Exercise.     The  area  of  a  square  is  16  sq.  in.     Find  the  radius 
of  the  inscribed  circle  and  also  the  radius  of  the  circumscribed  circle. 

Proposition  VI.     Problem 

734.  To  inscribe  a  regular  hexagon  in  a  circle, 

D 


Let  0  be  the  center  of  the  given  circle. 

Required  to  inscribe  a  regular  hexagon  in  the  circle. 
Draw  the  radius  OA.    Lay  off  the  chord  AB  =  OA.    Draw  OB. 
AOAB  \^  equilateral,  and  angle  0  contains  60°. 
.-.  the  arc  ^Z?  is  i  of  the  circumference,  and  the  chord  AB  is 
one  side  of  a  regular  hexagon. 

Complete  the  hexagon  ABCDEF.  q.e.f. 

735.  Corollary  I.  The  chords  joining  the  three  alternate 
vertices  form  an  inscribed  equilateral  triangle. 

736.  Corollary  II.  Tangents  drawn  at  the  vertices  of  the 
inscribed  hexagon  and  of  the  triangle  form  a  regular  circum- 
scribed heocagon  and  a  regular  circumscribed  triangle. 


BOOK   V  227 

737.  Corollary  III.  If  the  arcs  are  bisected  and  chords  and 
tangents  are  draivn  according  to  §  710,  regular  j^olygons  of  12, 
24,  48,  etc.,  sides  will  be  inscribed  in  and  circumscribed  about 
the  circle. 

738.  Exercise.  The  side  of  the  inscribed  equilateral  triangle  is 
i?V3,  and  its  area  is  |i?2\/3. 

739.  Exercise,  The  side  of  the  circumscribed  equilateral  triangle  is 
2  i?  V3,  and  its  area  is  3  i?2  V3. 

740.  Exercise.  The  side  of  a  regular  inscribed  hexagon  is  JK,  and  its 
area  is  f  ^'^Vs. 

741.  Exercise.  The  side  of  a  regular  circumscribed  hexagon  is 
|i?\/3,  and  its  area  is  2B'^y/Z. 

742.  Exercise.  The  area  of  a  regular  inscribed  hexagon  is  double 
that  of  an  equilateral  triangle  inscribed  in  the  same  circle.  [Show  this 
in  two  ways :  1st,  by  comparing  the  values  of  their  areas  as  derived  in 
§§  738  and  740 ;  2d,  by  a  geometrical  demonstration  using  the  figure  of 

§  734.] 

* 

743.  Exercise.  What  is  the  area  of  a  regular  hexagon  inscribed  in 
a  circle,  the  radius  of  which  is  4  in.  ? 

744.  Exercise.  The  area  of  a  regular  inscribed  hexagon  is  10  sq.  in. 
What  is  the  area  of  a  regular  hexagon  circumscribed  about  the  same 
circle  ? 

745.  Exercise.  The  area  of  an  equilateral  triangle  is  48  V3  sq.  ft. 
Find  the  radii  of  the  inscribed  and  of  the  circumscribed  circles. 

746.  Exercise.  The  area  of  a  regular  hexagon  is  54a2-v/3.  Find 
the  radii  of  the  inscribed  and  of  the  circumscribed  circles. 

747.  Exercise.  Show  that  the  circumscribed  equilateral  triangle  is  4 
times  the  inscribed  equilateral  triangle  ;  that  the  circumscribed  square 
is  2  times  the  inscribed  square  ;  and  that  the  circumscribed  regular  hexa- 
gon is  I  of  the  inscribed  regular  hexagon. 

748.  Exercise.  Divide  a  circumference  into  quadrants  by  the  use  of 
compasses  only. 

[Suggestion.  The  side  of  an  inscribed  square  is  the  altitude  of  an 
isosceles  triangle  whose  base  is  2  jR  and  one  of  whose  sides  is  B  V'3.  ] 


228  PLANE   GEOMETRY 

Proposition  YII.     Problem 
749.   To  inscribe  a  regular  decagon  in  a  circle. 


Let  0  be  the  center  of  the  given  circle. 
Required  to  inscribe  a  regular  decagon  in  the  circle. 
Draw  the  radius  OA.    Divide  it  into  extreme  and  mean  ratio, 
OB  being  the  greater  segment. 

Lay  off  ^C  =  OB.     Draw  BC  and  OC. 

By  definition  (Art.  551),  M  =  ^. 
^  ^    OB      BA 

AC      BA         ^^ 

AOAC  and  BAC  are  similar.     (§  495.) 
.-.  A  5^  C  is  isosceles,  and  ^G'  =  J5C.     (?) 
A  ^OC  is  isosceles.     (?) 

A1=Z3  +  Z0  (?)  or  Zl  =  2Zo.     (?) 
Aa  =  2Zo  (?)  and  Zaco  =  2Zo.     (?) 
Za+Zaco-{-Zo  =  1S0°.     (?) 
2Zo  +  2Zo-{-Zo  =  180°.     (?)     .-.  Zo  =  86°. 

.'.  the  arc  AC,  the  measure  of  Z  0,  contains  36°  of  arc,  and 
is  jij^  of  the  circumference. 

The  circumference  can  therefore  be  divided  into  ten  parts, 
each  equal  to  the  arc  AC,  and  the  chords  joining  the  points  of 
division  form  a  regular  inscribed  decagon.  q.e.f. 


BOOK   V  229 

750.  Corollary  I.  The  cliords  joining  the  alternate  vertices 
of  a  regular  inscribed  decagon  form  a  regular  inscribed  pentagon. 

751.  Corollary  II.  Tangents  drawn  at  the  vertices  of  the 
regular  inscribed  pentagon  and  decagon  form  a  regular  circwm- 
scribed  pentagon  and  a  regular  circumscribed  decagon. 

752.  Corollary  III.  If  the  arcs  are  bisected  and  chords 
and  tangents  are  drawn  according  to  §  710,  regular  inscribed 
and  circumscribed  polygons  of  20,  40,  80,  etc.,  sides  will  be  formed. 

753.  Exercise.  The  length  of  the  side  of  a  regular  inscribed  decagon 
is  Kv^ -!)»•• 

754.  Exercise.  Find  the  length  of  a  side  of  a  regular  inscribed  pen- 
tagon. [In  the  R.  A.  A  ADC  (see  the  figure  of  §  749),  ^O  is  the  side  of 
the  decagon,  and  AD  is  one  half  the  difference  between  the  radius  and  the 

side  of  the  decagon.]  VlO  -  2  V5 

Arts.   r. 

2 

755.  Exercise,  Show  that  the  sum  of  the  squares  described  on  the 
sides  of  a  regular  inscribed  decagon  and  of  a  regular  inscribed  hexagon 
equals  the  square  described  on  the  side  of  a  regular  inscribed  pentagon. 

[Represent  the  sides  of  the  pentagon,  hexagon,  and  decagon  by  p,  h, 
and  d,  respectively. 
In  the  figure  of  §  749, 

DC^  =  AC^  -  AD^, 

whence  p"^  =  n(P  -  h"^  +  2hd.  (1) 

By  §  551  ~  =  — ^  ,  whence  hd  =  h^  -  d^.  (2) 

d     h  —  d 

From  (1)  and  (2)  p'^  =  d^  +  h\ 

Give  also  to  algebraic  proof.] 

756.  Exercise.  What  is  the  length  of  the  side  of  a  regular  decagon 
inscribed  in  a  circle  having  a  diameter  4  in.  long  ? 

757.  Exercise.  If  the  side  of  a  regular  pentagon  is  2V5  in.,  show 
that  the  radius  of  the  circumscribed  circle  is  v  10  +  2  Vs  in. 


230  PLANE    GEOMETRY 

Proposition  VIII.     Problem 
758.    To  inscribe  a  regular  pentedecagon  in  a  circle. 


A 

Let  0  be  the  center  of  the  given  circle. 

Required  to  inscribe  a  regular  polygon  of  fifteen  sides  in  the 
circle. 

Lay  off  the  chord  AB  =  side  of  regular  inscribed  hexagon, 
and  the  chord  AC  =  side  of  regular  inscribed  decagon. 

The  arc  ^5  contains  60°,  (?)  and  the  arc  AC,  36°.  (?) 

.\  the  arc  J5(7  contains  24°  and  is  -^-^  of  the  circumference. 
The  circumference  can  therefore  be  divided  into  fifteen  parts, 
each  equal  to  BC;  and  the  chords  joining  the  points  of  division 
form  a  regular  inscribed  pentedecagon.  q.e.f. 

759.  Corollary  I.  Tangents  drawn  at  the  vertices  of  the 
inscribed,  pentedecagon  form  a  regidar  circumscribed  pentedecagon. 

760.  Corollary  II.  If  the  arcs  are  bisected,  and  chords  and 
tangents  are  drawn  as  described  in  §  710,  regular  inscribed  and 
circumscribed  polygons  of  30,  60,  120,  etc.,  sides  icill  be  formed. 

761.  Scholium.  In  Propositions  V.,  VL,  VII.,  and  VIII. 
we  have  seen  that  the  circumference  can  be  divided  into  the 
following  numbers  of  equal  parts  : 

2,  4,       8,       16    ...  2^ 

3,  6,     12,      24    ...      3x2" 
5,     10,     20,       40    ...      5x2" 

15,     30,     60,     120    ...    15  X  2'* 


n  being  any  positive 


BOOK  V  231 

The  mathematician  Gauss  has  shown  that  it  is  possible  to 
divide  the  circumference  into  2"  + 1  equal  parts,  n  being  a 
positive  integer  and  2"  + 1  a  prime  number. 

It  is  therefore  possible,  by  the  use  of  ruler  and  compasses, 
to  divide  the  circumference  into  2,  3,  5,  17,  257,  etc.,  equal 
parts. 

[An  elementary  explanation  of  the  division  of  the  circum- 
ference into  seventeen  equal  parts  is  given  in  Felix  Klein's 
"  Vortrage  liber  ausgewahlte  Fragerfder  Elementar  Geometric."] 

Proposition  IX.     Theorem 

762.  The  arc  of  a  circle  is  less  than  any  line  that 
envelops  it  and  has  the  same  extremities. 


Let  AMB  be  the  arc  of  circle  and  ASB  any  other  line  envelop- 
ing it  and  passing  through  A  and  B. 

To  Prove  AMB  <  ASB, 

Proof.  Of  all  the  lines  (a MB,  ASB,  etc.)  that  can  be  drawn 
through  A  and  B,  and  including  the  segment  or  area  AMB, 
there  must  be  one  of  minimum  length. 

ASB  cannot  be  the  minimum  line,  for  draw  the  tangent  CD 
to  the  arc  AMB. 

CD  <  CSD.      (?) 
ACDB  <ASB.      (?) 

The  same  can  be  shown  of  every  other  line  (except  AMB) 
passing  through  A  and  B  and  including  the  area  AMB. 

.-.  the  arc  AMB  is  the  minimum  line.  q.e.d. 


232  PLANE   GEOMETRY 

763.  CoKOLLARY  I.  The  circumference  of  a  circle  is  less  than 
the  perimeter  of  a  circumscribed  polygo7i  and  greater  than  the 
perimeter  of  an  inscribed  polygon. 

Proposition  X.     Theorem 

764.  If  the  number  of  sides   of  a  regular  inscribed 

polygon  is  indefinitely  increased,  its  apothem  approaches 

the  radius  as  a  limit.     ^ 

A 


Let  AB  be  the  side  of  a  regular  inscribed  polygon  and  OC  be 
its  apothem. 

To  Prove  that  OC  approaches  the  radius  as  its  limit  when 
the  number  of  sides  is  indefinitely  increased. 

Proof.  OA>OC.     (?) 

OA—OC<AC.      (?)        .-.    OA  —  OC  <AB. 

By  increasing  the  number  of  sides  AB  can  be  made  as  small 
as  we  please,  but  not  equal  to  zero.  AB  consequently  ap- 
proaches zero  as  a  limit,  and  since  OA—  OC  <AB,  OA  —  OC 
approaches  zero  as  its  limit;  and  OC  approaches  OA  as  its 
limit.  Q.E.D. 

765.  Corollary.  If  the  number  of  sides  of  a  regular 
circumscribed  polygon  is  indefinitely  increased,  the  distance 
from  a  vertex  to  the  center  of  the  circle  approaches  the  radius 
as  a  limit. 

[Proof  similar  to  §  764.] 


BOOK    V 


233 


Proposition  XI.     Theorem 

766.  If  a  regular  polygon  is  inscribed  in  or  circum- 
scribed about  a  circle  and  the  number  of  its  sides  is 
indefinitely  increased, 

I.  The  perimeter  of  the  polygon  approaches  the  cir- 
cumference as  its  limit. 

II.  The  area  of  the  polygon  approaches  the  area  of 
the  circle  as  its  limit. 


Let  AB  be  the  side  of  a  regular  circumscribed  polygon,  and 
CB  (parallel  to  AB)  be  the  side  of  a  similar  inscribed  polygon. 

I.  To  Prove  that  the  perimeters  of  the  polygons  approach 
the  circumference  of  the  circle  as  a  limit  when  the  number  of 
sides  is  indefinitely  increased. 

Proof.     Draw  OA^  OB,  and  OE. 
OA  passes  through  C  and  OB  through  D.     (?) 
Let  P  and  p  stand  for  the  perimeters  of  the  circumscribed 
and  inscribed  polygons  respectively. 


OE 
OF 


P  —P_OE  —  OF 
P      ~        OE 


(?) 


(?) 


or 


p=^-=--(OE—  OF). 
OE 


234  PLANE   GEOMETRY 

As  shown  in  the  preceding  proposition,  OE  —  OF  can  be 
made  as  small  as  we  please,  though  not  equal  to  zero;   and 

T*  ,  p 

since  — ;   does  not  increase,  — -{OE  —  OF),  or  its  equal  P  —p, 
OE  OE 

can  be  decreased  at  pleasure. 

Since  P  is  always  greater  than  the  circumference,  and  p  is 
always  less  than  the  circumference,  the  difference  between  the 
circumference  and  either  perimeter  is  less  than  the  difference 
P  -^p,  and  can  consequently  be  made  as  small  as  we  please,  but 
not  equal  to  zero. 

The  circumference  is  therefore  the  common  limit  of  the  two 
perimeters  as  the  number  of  sides  is  indefinitely  increased. 

Q.E.D. 

II.  To  Prove  that  the  areas  of  the  polygons  approach  the 
area  of  the  circle  as  a  limit,  when  the  number  of  sides  is 
indefinitely  increased. 

Proof.  Let  S  and  s  stand  for  the  areas  of  the  circumscribed 
and  inscribed  polygons  respectively. 

^=^.       (9) 

s      of'      ^'^ 
S  —  S  ^  0E'~  of'  _  CF^^       ,^. 

c,  — -2  2*         ^'z 

S  OE  OE 

,S-S=:-^^(Cf') 

OE' 

As  the  number  of  sides  is  indefinitely  increased,  CD  ap- 
proaches zero  as  a  limit,  as  does  also  CF,  and  consequently 

cf'. 

[The  remainder  of  the  proof  is  similar  to  that  of  Case  I.  of 
this  proposition.]  q.e.d. 

767.  Exercise.  If,  as  is  shown  in  §  706,  the  difference  between  P  and 
p  can  be  made  as  small  as  we  please,  why  is  not  p  the  limit  of  P?  (See 
definition  of  limit.) 


BOOK    V 


235 


Pkopositiox  XII.     Problem 
768.   Given  the  perUneters  of  a  regular  inscribed  polygon 
and  of  a  similar  circumscribed  polygon,  to  find  the  perim- 
eters of  regular  inscribed  and  circumscribed  polygons 
of  double  the  number  of  sides. 

G 


Let  ^17?  bo  a  side  of  a  regular  inscribed  polygon  of  n  sides, 
CD  (parallel  to  ^£)  a  side  of  a  regular  circumscribed  polygon 
of  n  sides, 

AE  a  side  of  a  regular  inscribed  polygon  of  2  n  sides, 
FG  a  side  of  a  regular  circumscribed  polygon  oi2n  sides. 
Required  to  find  the  perimeters  of  the  inscribed  and  circum- 
scribed polygons  of  2  n  sides. 

Call  the  perimeter  of  the  inscribed  polygon  of  n  sides  p, 
the  perimeter  of  the  circumscribed  polygon  of  n  sides  P, 
the  perimeter  of  the  inscribed  polygon  of  2  n  sides  |)', 
the  perimeter  of  the  circumscribed  polygon  of  2n  sides  P\ 


Then      AB 


=  ^  and^^=/-. 
n  Z  n 

P' 
AE  =  -~^.      FG 

2n 

P       OF 
P  -\-P       CF  -\-  FF 


2p 


P'=^ 


2FF 
2p  X  P 
P+P 


CD  : 

2n 

CF 

FF 
_  CF 
~  FG 


P  ,  P 

—  and  CE  =  ^r—i 
n  2n 


(§  502.) 

_p_ 
"  p'' 


(I-) 


236 


PLANE   GEOMETRY 


Prove  A IFE  and  AEH  similar. 


whence 


^  =  4  and 
AE      P' 

P 

'  P' 


All _  IE 
AE~  FE 
IE  ^  p\ 
FE~  F' 


(?) 


~  and  P'  =  Vp  X  F'.     (II.) 


Since  p  and  F  are  given,  Formula  I.  gives  the  value  of  p'; 
then  from  Formula  II.  the  value  of  p'  can  be  derived.        q.e.f. 

769.  Exercise.  The  side  of  an  inscribed  square  is  3\/2  and  the  side 
of  a  circumscribed  square  is  6.  Find  the  sides  of  regular  octagons  in- 
scribed in  and  circumscribed  about  the  same  circle. 

770.  Exercise.  Eind  the  perimeters  of  regular  dodecagons  (12-sided 
polygons)  inscribed  in  and  circumscribed  about  a  circle  having  a  diameter 
12  in.  long. 

Proposition  XIII.     Theorem 

771.  The  area  of  a  regular  poly g on  is  equal  to  one  half 
the  product  of  its  perimeter  and  apothem. 


Let  ABCDEF  be  a  regular  polygon. 

To  Prove  that  its  area  is  equivalent  to  one  half  the  product  of 
its  perimeter  and  apothem. 

Suggestion.    The  altitude  of  each  A  is  the  apothem,  and  the  polygon 
is  equivalent  to  the  sum  of  the  triangles. 


BOOK  V  237 

772.  CoROLLARr.  The  area  of  any  circumscribed  polygon  is 
equal  to  one  half  the  product  of  its  perimeter  and  the  radius  of 
its  inscribed  circle. 

773.  Exercise.  The  perimeter  of  a  polygon  circumscribed  about 
a  circle  having  a  5  ft.  radius,  is  32  ft.     What  is  its  area  ? 

774.  Exercise.  The  side  of  a  regular  hexagon  is  6  in.  Find  its 
area.     \_SuggesUon.     First  find  its  apothem.] 

Proposition  XIV.     Theorem 

775.  The  area  of  a  circle  is  equal  to  one  half  the 
product  of  its  circumference  and  radius. 

A 


C 
Let  xyz  be  any  circle. 

To  Prove  area  xyz  =  |-  circumference  x  radius. 
Proof.     Circumscribe  a  regular  polygon  ABO  about  the  circle 
^y^'  Area  ABC  =  ^  perimeter  x  apothem.     (?) 

If  the  number  of  sides  of  the  polygon  is  increased,  the  area 
changes  as  does  also  the  perimeter,  and  yet  the  area  is  always 
equal  to  ^  perimeter  x  apothem.  So  the  two  members  of  the 
above  equation  may  be  regarded  as  two  variables  that  are 
always  equal.  Since  each  is  approaching  a  limit,  their  limits 
must  be  equal.     [§  341.] 

The  limit  of  area  ABC  =  area  of  circle.  (?) 

The  limit  of  the  perimeter  =  circumference.     (?) 
The  apothem  is  constant  and  equals  the  radius. 
.*.  area  xyz  =  ^  circumference  x  radius.  q.e.d. 


238  PLANE   GEOMETRY 

776.    Corollary.     The  area  of  a  sector  is  equal  to  one  half 
the  product  of  its  arc  and  radius. 

To  Prove  area  AOB  =  ^  AB  x  R.  C_ 

Proof.     Construct  the  quadrant  COD.         /^ 
sector^O-B     ^5 /^  o^r-\  / 


or 


sector  COB 

CB  ^ 

sector  AOB 

AB 

4  sector  COB 

4  CB 

sector  AOB 

AB 

circle 

circumf. 

sector  A  OB 

_       ^^ 

..  sector  ^0J5  =  i  ^5  X  i^. 
\  circumf.  X  B      circumf. 

777.  Exercise.  The  radius  of  a  circle  is  100  ft.  and  its  circumference 
is  628.32  ft.     Find  its  area. 

778.  Exercise.  The  area  of  a  sector  is  68  sq.  in.,  and  its  radius  is 
8  in.     How  long  is  its  arc  ? 

779.  Exercise.  The  area  of  a  circle  is  100  sq.  ft.  The  area  of  a 
sector  of  this  circle  is  12|  sq.  ft.  How  many  degrees  in  the  arc  of  the 
sector  ? 

780.  Exercise.  The  radius  of  a  circle  is  10  ft.  Eind  the  area  of  a 
segment  whose  arc  contains  60°. 

Suggestion.  Find  the  area  of  the  sector  having  arc  =  60°.  Subtract 
the  area  of  the  triangle  formed  by  the  chord  and  the  radii  from  the  area 
of  the  sector. 

781.  Exercise.  The  circumference  of  a  circle  is  94.248  ft.  The  side 
of  an  inscribed  equilateral  triangle  is  15  VS  ft.  Find  the  area  of  the 
circle. 

782.  Exercise.  The  area  of  a  circle  is  314.16  sq.  in.  The  perimeter 
of  a  regular  inscribed  hexagon  is  60  in.  Find  the  circumference  of  the 
circle. 

783.  Exercise.  Find  the  area  of  the  part  of  the  circle  of  §  782 
lying  between  its  circumference  and  the  perimeter  of  a  regular  hexagon 
inscribed  in  the  circle. 


BOOK   V  239 

Proposition  XV.     Theorem 

784.  TJie  circuTrbferences  of  two  circles  are  to  each 
other  as  their  radii,  and  the  circles  are  to  each  other  as 
the  squares  of  their  radii. 


Let  A  and  B  be  two  circles  and  R  and  r  be  their  radii. 

_    _.  circumf.  A      R 

To  Prove =  -  • 

circumi.   B      r 

Proof.  Inscribe  similar  regular  polygons  in  the  two  circles. 
Let  P  and  p  denote  the  perimeters  of  these  polygons. 

P^R      (.^    or     ^  =  TP-     (1) 
p      r      ^  R      r     ^  ' 

As  the  number  of  sides  is  indefinitely  increased,  P  and  p  ap- 
proach circumference  A  and  circumference  B  respectively  as 
their  limits.     (?) 

The  members  of  equation  (1)  may  therefore  be  regarded  as 
two  variables  that  are  always  equal,  and  since  each  is  approach- 
ing a  limit,  their  limits  are  equal.     (?) 

circumf.   A  _  Circumf.  B 
R  r 

circumf.   A  _R^ 
circumf.   B      r 

Similarly,  show  that  ^!^^|^  ^  =  ^.  Q.e.d. 

circle  B      ?- 


or 


240  PLANE   GEOMETRY 

785.  Corollary  I.  The  circuinferences  of  two  circles  are  to 
each  other  as  their  diameters,  and  the  circles  are  to  each  other  as 
the  squares  of  their  diameters. 

786.  Corollary  II.  The  ratio  of  the  circumference  of  a 
circle  to  its  diameter  is  constant ;  that  is,  it  is  the  same  for  all 
circles. 

circumf.  A      diam.  A 


By  §  785, 


OP 


circumf.  B      diam.  b' 
circumf.  A      circumf.  B 


diam .  A  diam.  B 

Tlie  value  of  this  constant  is  denoted  by  the  Greek  letter  n. 

mi,  circumf.  A 

Thus,  -— =  TT. 

diam.  A 

Whence  circumf.  A  =7r  diam.  A. 

i.e.   Tlie  circumference  of  a  circle  is  tt  times  its  diameter. 

If,  in  the  formula  for  the  area  of  a  circle, 

area  =  ^  circumf.  x  R, 

the  value  of  the  circumference  just  derived  is  substituted,  we 
obtain 

area  =  ttR^. 

i.e.   The  area  of  a  circle  is  ir  times  the  square  of  its  radius. 

787.  Definition.  Similar  arcs  are  arcs  that  subtend  equal 
angles  at  the  center. 

Since  the  intercepted  arcs  are  the  measures  of  the  angles 
at  the  center,  similar  arcs  contain  the  same  number  of  degrees 
of  arc,  and  are  consequently  like  parts  of  their  circumferences. 

Similar  sectors  are  sectors  the  radii  of  v/hich  include  equal 
angles,  or  intercept  similar  arcs. 

Similar  segments  are  segments  whose  arcs  are  similar. 


BOOK    V  241 

788.  Corollary  III.     Similar  arcs  are  to  each  other  as  their 
radii.     [See  definition.] 

789.  Corollary  IV.     Similar  sectors  are  to  each  other  as 
the  squares  of  their  radii.     [§§  776  and  788.] 

790.  Corollary  V.     Similar  segments  are  to  each  other  as 
the  squares  of  their  radii. 

791.  Exercise.     The  circumferences  of  two  circles  are  942.48  ft.  and 
157.08  ft.  respectively. 

The  diameter  of  the  first  is  300  ft.     Find  the  diameter  of  the  second. 

792.  Exercise.     What  is  the  ratio  of  the  areas  of  the  two  circles  of 
the  preceding  exercise  ? 

793.  Exercise.     How  many  units  in  the  radius  of  a  circle,  the  area 
and  circumference  of  which  can  be  expressed  by  the  same  number  ? 


Proposition  XVI.     Problem 
794.   To  find  an  approximate  value  of  tt. 

The  perimeter  of  a  circumscribed  square  (see  §  729)  is  4  i) 
(p  =  diameter). 

The  perimeter  of  an  inscribed  square  is  2  V2  Z) =2.8284271  D. 
Substituting  4  7)  for  P  and  2.8284271  D  for  p  in  the  formulas 

p'  =  ^  ^  ^  (1)  and  p'  —  ^p  X  P'  (2),  we  get  P'  or  the  perime- 
ter of  the  circumscribed  octagon  ==  3.3137085  D,  and  i?'  or  the 
perimeter  of  the  inscribed  octagon  =  3.0614675  B. 

Substituting  3.3137085  B  for  P  and  3.0614675  B  fori?  in  for- 
mulas (1)  and  (2),  we  obtain  values  for  the  perimeters  of  the 
circumscribed  and  the  inscribed  polygons  of  sixteen  sides. 

Substituting  these  values,  the  perimeters  of  polygons  of 
thirty -two  sides  are  obtained. 

SANDERS'    GEOM. 16 


242  PLANE    GEOMETRY 

Continuing  in  this  way,  the  following  table  is  formed : 


Number  of 

Perimeter  of 

Perimeter  of 

SIDES. 

CIRCUMSCRIBED  POLYGON. 

INSCRIBED  POLYGON. 

4 

4.0000000 1) 

2.8284271 D 

8 

3.3137085  i) 

3.0614675  2) 

16 

3.1825979  D 

3.1214452  2) 

32 

3.1517249  2)  * 

3.1365485  D 

64 

3.1441184  D 

3.1403312  2) 

128 

3.1422236  D 

3.1412773  2) 

256 

3.1417504  D 

3.1415138  2) 

512 

3.1416321  D 

3.1415729  2) 

1024 

3.1416025  z> 

3.1415877  2) 

2048 

3.14159512) 

3.1415914  2) 

4096 

3.1415933  2) 

3.1415923  2) 

8192 

3.1415928  D 

3.1415926  2)  . 

The  circumference  of  the  circle  therefore  lies  between 
3.14159262)  and  3.1415928  2). 

For  ordinary  accuracy  the  value  of  tt  is  taken  as  3.1416. 

Note.  —  The  vahie  of  tt  has  been  carried  out  over  seven  hundred 
decimal  places.  [See  article  on  "Squaring  the  Circle"  in  the  Encyclo- 
psedia  Britannica.  ] 

The  value  of  ir  to  thirty-five  decimal  places  is 

3.14159265358979323846264338327950288. 

By  higher  mathematics,  the  diameter  and  circumference  of 
the  circle  have  been  shown  to  be  incommensurable,  so  no  exact 
expression  for  their  ratio  can  be  obtained. 

795.  Exercise.  The  radius  of  a  circle  is  10  in.  Find  its  circum- 
ference and  its  area. 

796.  Exercise.  The  area  of  a  circle  is  7854  sq.  ft..  Find  its  cir- 
'  cumf  erence. 


BOOK  V  243 

797.  Exercise.  The  circumference  of  a  circle  is  50  in.  What  is  its 
area  ? 

798.  Exercise.  The  radius  of  a  circle  is  60  ft.  What  is  the  area  of 
a  sector  whose  arc  contains  40°  ? 

799.  Exercise.  The  radius  of  a  circle  is  10  ft.  The  area  of  a  sector 
of  that  circle  is  120  sq.  ft.     What  is  its  arc  in  degrees  ? 

EXERCISES 

1.  In  a  regular  polygon  of  n  sides,  diagonals  are  drawn  from  one 
vertex.     What  angles  do  they  make  with  each  other  ? 

2.  Show  that  the  altitude  of  an  inscribed  equilateral  triangle  is  |  of 
the  diameter,  and  that  the  altitude  of  a  circumscribed  equilateral  triangle 
is  3  times  the  radius. 

3.  The  radii  of  two  circles  are  4  in.  and  6  in.  respectively.  How  do 
their  areas  compare  ? 

4.  Find  the  area  of  the  ring  between  the  circumferences  of  two  con- 
centric circles  the  radii  of  which  are  a  and  h  respectively. 

5.  The  area  of  a  regular  inscribed  hexagon  is  a  mean  proportional 
between  the  areas  of  the  inscribed  and  the  circumscribed  equilateral  tri- 
angles.    [See  Ex.  to  Prop.  6.  ] 

6.  The  diagonals  joining  the  alternate  vertices 
of  a  regular  hexagon  form  by  their  intersection  a 
regular  hexagon  having  an  area  one  third  of  that 
of  the  original  hexagon. 

7.  Eind  the  area  of  the  six-pointed  star  in  the 
figure  of  Exercise  6  in  terms  of  the  radius  of  the 
circle. 

8.  Erom  any  point  within  a  regular  polygon 
of  n  sides,  perpendiculars  are  drawn  to  the  sides. 

Prove  that  the  sum   of   these    perpendiculars    is  equal  to  n  times  the 
apothem  of  the  polygon. 

[Join  the  point  with  the  vertices  and  obtain  an  expression  for  the  area 
of  the  polygon.  Compare  this  with  the  expression  for  the  area  obtained 
from§  771.] 

9.  Construct  a  circle  that  shall  be  double  a  given  circle  (§  784). 

10.  Construct  a  circle  that  shall  be  one  half  a  given  circle. 

11.  Construct  a  circle  equivalent  to  the  sum  of  two  given  circles  ;  also 
one  equivalent  to  their  difference.     [§  646.] 


244 


PLANE   GEOMETRY 


12.  If  two  circles  are  concentric,  show  that  the  area  of  the  ring 
between  their  circumferences  is  equal  to  the  area  of  a  circle  having  for 
its  diameter  a  chord  of  the  larger  circle  that  is  tangent  to  the  smaller. 

13.  Find  the  area  of  the  sector  of  a  circle  intercepting  an  arc  of  50°, 
the  radius  of  the  circle  being  10  ft.     [§  776.] 

14.  The  radius  of  a  circle  is  20  ft.  What  is  the  angle  of  a  sector 
having  an  area  of  300  sq.  ft.  ? 

15.  The  radius  of  a  circle  is  20  ft.,  and  the  area  of  a  sector  of  the 
circle  is  300  sq.  ft.  Find  the  area  of  a  similar  sector  in  a  circle  having 
a  radius  50  ft.  long. 

16.  What  is  the  radius  of  a  circle  having  an  area  equal  to  16  times  the 
area  of  a  circle  with  a  radius  5  ft.  long  ? 

17.  Find  the  area  of  a  circle  circumscribed  about  a  square  having  an 
area  of  600  sq.  ft.     [§  729.J 

18.  Show  that  the  area  of  a  circumscribed  equilateral  triangle  is  greater 
than  that  of  a  square  circumscribed  about  the  same  circle. 

19.  Four  circles,  each  with  a  radius  5  ft.  long,  have  their  centers  at 
the  vertices  of  a  square,  and  are  tangent.  Find  the  area  of  a  circle  tan- 
gent to  all  of  them. 

20.  How  many  degrees  in  the  arc,  the  length  of  which  is  equal  to  the 
radius  of  the  circle  ? 

21.  A  circle  is  circumscribed  about  the  right- 
angled  triangle  ABC.  Semicircles  are  described 
on  the  two  legs  as  diameters.  Prove  that  the 
sum  of  the  crescents  AD  BE  and  BFCG  is  equiv- 
alent to  the  triangle  ABC. 

22.  The  radius  of  a  regular  inscribed  polygon 
is  a  mean  proportional  between  its  apothem  and 
the  radius  of  a  similar  circumscribed  polygon. 

23.  If  the  bisectors  pf  the  angles  of  a  polygon 
meet  in  a  point,  a  circle  can  be  inscribed  in  the 
polygon. 

24.  The  diagonals  of  a  regular  pentagon  form 
by  their  intersection  a  second  regular  pentagon. 

25.  Any  two  diagonals  of  a  regular  pentagon 
not  drawn  from  a  common  vertex  divide  each 
other  into  extreme  and  mean  ratio.  [A  ABC 
and  CfD  are  similar.] 

26.  Divide  an  angle  of  an  equilateral  triangle  into  five  equal  parts. 


BOOK    V 


245 


27.  If  two  angles  at  the  centers  of  unequal  circles  are  subtended  by 
arcs  of  equal  length,  the  angles  are  inversely  proportional  to  the  radii  of 
the  circles. 

28.  The  apothem  of  a  regular  inscribed  pentagon  is  equal  to  one  half 
the  sum  of  the  radius  of  the  circle  and  a  side  of  a  regular  inscribed 
decagon. 

29.  If  two  chords  of  a  given  circle  intersect  each  other  at  right  angles, 
and  on  the  four  segments  of  the  chords  as  diameters,  circles  are  de- 
scribed, the  sum  of  the  four  circles  is  equivalent  to  the  given  circle.  [Ex. 
34,  page  217.] 

30.  Divide  a  circle  into  three  equivalent 
parts  by  concentric  circles  (§  784). 

31.  The  radius  of  a  given  circle  ABD  is  10 
ft.  Find  the  areas  of  the  two  segments  BCA 
and  BDA  into  which  the  circle  is  divided  by  a 
chord  AB  equal  in  length  to  the  radius.  [Sub- 
tract area  of  A  from  area  of  sector.  ] 

32.  Find  the  radius  of  a  circle  that  is 
doubled  in  area  by  increasing  its  radius  one 
foot. 

33.  On  the  sides  of  a  square  as  diameters, 
four  semicircles  are  described  within  the 
square,  forming  four  leaves.  If  the  side  of  the 
square  is  a,  find  the  area  of  the  leaves. 

34.  In  a  given  equilateral  triangle  inscribe 
three  equal  circles  tangent  to  each  other  and  to 
the  sides  of  the  triangle. 

35.  In  a  given  circle  inscribe  three  equal 
circles  tangent  to  each  other  and  to  the  given 
circle. 

36.  In  the  circle  ABCD,  the  diameters  AC 
and  BD  are  at  right  angles  to  each  other.  With 
E,  the  middle  point  of  00,  as  a  center,  and  BB 
as  a  radius,  the  arc  BF  is  described.  Prove 
that  the  radius  OA  i.^  divided  into  extreme 
and  mean  ratio  at  F. 

[Describe  arc  OG  with  E  as  center,  and  arc 
G^^with  £  as  center.] 


246  PLANE    GEOMETRY 

37.  The  diameter  AB  of  a  given  circle  is  divided  into  tv/o  segments, 
AC  and  CB.  On  each  segment  as  a  diameter  a  semicircle  is  described, 
but  on  opposite  sides  of  the  diameter.  Prove  that 
the  sum  of  the  two  semi-circumferences  described 
is  equal  to  the  semi-circumference  of  the  given 
circle,  and  that  the  line  they  form  divides  the 
given  circle  into  parts  that  are  to  each  other  as 
the  segments  of  the  diameter. 

38.  If  a  given  square  is  divided  into  four  equal 
squares,  and  a  circle  is  inscribed  in  each  of  the  small  squares  and  also  in 
the  given  square,  prove  that  the  sum  of  the  four  small  circles  is  equiva- 
lent to  the  circle  inscribed  in  the  given  square. 

39.  If  a  regular  polygon  of  n  sides  be  circumscribed  about  a  circle,  the 
sum  of  the  perpendiculars  from  the  points  of  contact  to  any  tangent  to 
the  circle  is  equal  to  n  times  the  radius. 

[If  J.,  B,  C,  Z),  etc.,  are  the  points  of  contact  of  the  polygon  and  P  the 
point  at  which  the  tangent  is  drawn,  the  sum  of  the  Js  from  A,  B,  etc., 
on  tangent  at  P  =  sum  of  Js  from  P  to  tangents  drawn  at  A,  B,  etc. ;  and 
this  by  Ex.  8  =  wP.] 

40.  The  sum  of  the  perpendiculars  from  the  vertices  of  a  regular 
inscribed  polygon  to  any  line  without  the  circle  is  equal  to  n  times  the 
perpendicular  from  the  center  of  the  circle  to  the  line. 

[Draw  a  tangent  to  the  O  parallel  to  the  given  line,  and  then  use  Ex.  39.] 

41.  The  sum  of  the  squares  of  the  lines  drawn  from  any  point  in  the  cir- 
cumference to  the  vertices  of  a  regular  inscribed  polygon  is  equal  to  2  nE^. 

[Using  notation  of  Ex.  39,  show  that  the  square 
of  the  line  from  the  given  point  P  to  each  vertex 
=  2  P  times  the  ±  from  the  vertex  to  a  tangent 
at  P.     Add  these  equations  and  use  Ex.  39.] 

42.  A  crescent-shaped  region  is  bounded  by 
a  semi-circumference  of  radius  a,  and  another 
circular  arc  whose  center  lies  on  tlie  semi-circum- 
ference produced.  Find  the  area  and  the  perim- 
eter of  the  region. 

[Show  that  the  arc  is  a  quadrant  in  a  O  with 
radius  =  a  \/2.] 

43.  Three  points  divide  a  circumference  into 
equal  parts.  Through  each  pair  of  these  points 
an  arc   of  a  circle  is  described  tangent  to  the 


BOOK   V  247 

radii  drawn  to  the   points    and    lying  wholly  within  the  circle.     Find 
the  perimeter    of   the  figure  thus    formed,    and   show   that   its   area  is 
3  ( V3  —  I  tt)  a2^  where  a  denotes  the  radius  of 
the  circle. 

[Show  that  each  arc  is  ^  of  a  circumference 
with  radius  a  VS.] 

44.  Three  radii  are  drawn  in  a  circle  of 
radius  2  a,  so  as  to  divide  the  circumference 
into  three  equal  parts ;  and,  with  the  middle 
of  these  radii  as  centers,  arcs  are  drawn,  each 
with  the  radius  a,  so  as  to  form  a  closed  figure 

(trefoil).     Show  that  the  length  of  the  perimeter  of  the  trefoil  is  equal  to 
that  of  the  circle,  and  find  its  area. 


RETURN  TO  the  circulation  desk  of  any 
University  of  California  Library 
or  to  the 

NORTHERN  REGIONAL  LIBRARY  FACILITY 
BIdg.  400,  Richmond  Field  Station 
University  of  California 
Richmond,  CA  94804-4698 


ALL  BOOKS  MAY  BE  RECALLED  AFTER  7  DAYS 
2-month  loans  may  be  renewed  by  calllna 

(510)642-6753 
1-year  loans  may  be  recharged  by  bringing  books 

Renewals  and  recharges  may  be  made  4  days 
prior  to  due  date 


DUE  AS  STAMPED  BELOW 


'mi 


9  1993 


YB   17302 


A-/^'^  '■■J'>::.S'^r%k  ■  ■  •  ■;•<  ' ^  ■; ,>':.:■  h-kk 


WMii 


